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REESE    LIBRARY 

OF    THR 

UNIVERSITY   OF  CALIFORNIA. 

Received— 


CUS^  • 
Accessions  No..'~F&.&..&      Shelf  No 


V 


THE 


PROGRESSIVE 


PRACTICAL  ARITHMETIC, 


CONTAINING 

THE  THEORY  OF  NUMBERS,  IX  CONNECTION  WITH  CONCISE  ANALYTIC 

AND  SYNTHETIC  METHODS  OF   SOLUTION,  AND  DESIGNED 

AS  A  COMPLETE  TEXT-BOOK  ON  THIS  SCIENCE, 

FOR 

COMMON  SCHOOLS  AND  ACADEMIES. 


BY 

HORATIO   N.  ROBINSON,  LL.D., 

AUTHOR    OF    A    WORK    ON     ALGEBJft*J '  ^^METiiTIS^SfcSlf^OMETRY,    SURVEYING    ATO 
NAVIGATION,    ASTRONMjfc.'  fUirBBBNTIAL '  JUffl  l&TJRpR^L    CALCULUS,  ETC. 


UNIVERSITY 


IVISON  &  PHINNEY,  48  &  50  WALKER  ST. 
CHICAGO:    S.  C.  GRIGGS    &   CO. 

CINCINNATI :     MOORE,    WILSTACH,    KEYS    &    CO.       PHILADELPHIA  :    SOWER,   BARNES    *    CO. 

BOSTON:    BROWN,     TAGGARD    AND    CHASE.        ST.    LOUIS:     KEITH     AND     WOODS. 

NEW  ORLEANS  :  BLOOMFIELD,  STEELE  &  CO.     BUFFALO :  PHINNEY  &  CO. 

1859. 


Entered,  according  to  Act  of  Congress,  in  the  year  1858,  by 
HORATIO  N.  ROBINSON,   LL.D., 

la  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Northern 
District  of  New  Yorl 


ELECTROTYPE!)    AT    THE 
BOSTON     STEREOTYPE     FOUNDRY. 


UNIVERSITY 


PROGRESS  and  improvement  characterize  almost  every  art  and 
science ;  and  within  the  last  few  years  the  science  of  Arithmetic  has 
received  many  important  additions  and  improvements,  which  have 
appeared  from  time  to  time  successively  in  the  different  treatises  pub- 
lished upon  this  subject. 

In  the  preparation  of  this  work  it  has  been  the  author's  aim  to  com- 
bine, and  to  present  hi  one  harmonious  whole,  all  these  modern  im- 
provements, as  well  as  to  introduce  some  new  methods  and  practical 
operations  not  found  in  other  works  of  the  same  grade ;  in  short,  to 
present  the  subject  of  Arithmetic  to  the  pupil  more  as  a  science  than 
an  art ;  to  teach  him  methods  of  thought,  and  how  to  reason,  rather 
than  what  to  do;  to  give  unity,  system,  and  practical  utility  to  the 
science  and  art  of  computation. 

The  author  believes  that  both  teacher  and  pupil  should  have  the 
privilege,  as  well  as  the  benefit,  of  performing  at  least  a  part  of  the 
thinking  and  the  labor  necessary  to  the  study  of  Arithmetic ;  hence 
the  present  work  has  not  "been  encumbered  with  the  multiplicity  of 
"notes,"  "suggestions,"  and  superfluous  operations  so  common  to 
most  Practical  Arithmetics  of  the  present  day,  and  which  prevent  the 
cultivation  of  that  self-reliance,  that  clearness  of  thought,  and  that 
vigor  of  intellect,  which  always  characterize  the  truly  educated  mind. 

The  author  claims  for  this  treatise  improvement  upon,  if  not  superi- 
ority over,  others  of  the  kind  in  the  following  particulars,  viz. :  In 
the  mechanical  and  typographical  style  of  the  tcork ;  the  open  and 
attractive  page ;  the  progressive  and  scientific  arrangement  of  the 
subjects ;  clearness  and  conciseness  of  definitions ;  fullness  and  accu- 
racy in  the  new  and  improved  methods  of  operations  and  analyses; 
brevity  and  perspicuity  of  rules;  and  in  the  very  large  number  of 

(iii) 


IV  PREFACE. 

examples  prepared  and  arranged  with  special  reference  to  their  prac- 
tical utility,  and  their  adaptation  to  the  real  business  of  active  life. 
The  answers  to  a  part  of  the  examples  have  been  omitted,  that  the 
learner  may  acquire  the  discipline  resulting  from  verifying  the  opera- 
tions. 

Particular  attention  is  invited  to  improvements  in  the  subjects  of 
Common  Divisors,  Multiples,  Tractions,  Percentage,  Interest,  Propor- 
tion, Analysis,  Alligation,  and  the  Roots,  as  it  is  believed  these 
articles  contain  some  practical  features  not  common  to  other  authors 
upon  these  subjects. 

It  is  not  claimed  that  this  is  a  perfect  work,  for  perfection  is  impossi- 
ble ;  but  no  effort  has  been  spared  to  present  a  clear,  scientific,  com- 
prehensive, and  complete  system,  sufficiently  full  for  the  business  man 
and  the  scholar ;  not  encumbered  with  unnecessary  theories,  and  yet 
combining  and  systemizing  real  improvements  of  a  practical  and 
useful  nature.  How  nearly  this  end  has  been  attained  the  intelligent 
and  experienced  teacher  and  educator  must  determine. 

In  conclusion,  the  author  takes  this  opportunity  of  acknowledging 
his  obligations  to  those  teachers  from  whom  he  has  received  valuable 
hints  and  suggestions,  many  of  which  have  been  incorporated  into 
this  work.  He  desires  also  to  make  particular  acknowledgment  for 
the  valuable  services  rendered  in  the  preparation  of  this  work  by  Mr. 
D.  W.  Fish,  of  Rochester,  N.  Y.,  who  mostly  prepared  the  Primary 
and  Intellectual  Arithmetics  of  this  series,  and  Mr.  J.  H.  French, 
of  Syracuse,  N.  Y.,  Superintendent  of  the  New  York  State  Map  and 
Atlas  Survey.  These  gentlemen  have  had  long  and  successful  experi- 
ence as  teachers,  and  to  them  the  public  are  indebted  for  the  excellent 
plan  and  superior  arrangement  of  this  treatise,  and  for  many  of  the 
new  and  practical  features  which  it  contains. 


CONTENTS. 


SIMPLE    NUMBERS. 

Page 

Definitions, 9 

Roman  Notation, 10 

Table  of  Roman  Notation, , .  11 

Arabic  Notation, 12 

Numeration  Table, 17 

Laws  and  Rules  for  Notation  and  Numeration, 18 

Addition, 22 

Subtraction 32 

Multiplication, 41 

Contractions, 48 

Division 53 

Contractions, 63 

Applications  of  preceding  Rules, 67 

General  Principles  of  Division, 72 

Exact  Divisors, 74 

Prime  Numbers, 75 

Factoring  Numbers, 75 

Cancellation, 77 

Greatest  Common  Divisor, 81 

Multiples,  87 

Classification  of  Numbers, 92 

COMMON    FRACTIONS. 

Definitions,  &c., 94 

General  Principles  of  Fractions, 97 

Reduction  of  Fractions, , 97 

*  (v) 


VI  CONTENTS. 

Page 

Addition  of  Fractions, 104 

Subtraction  of  Fractions, 106 

Multiplication  of  Fractions, 109 

Division  of  Fractions, 114 

Promiscuous  Examples, 120 

DECIMALS. 

Decimal  Notation  and  Numeration, 124 

Reduction  of  Decimals, 129 

Addition  of  Decimals, 132 

Subtraction  of  Decimals, 134 

Multiplication  of  Decimals, 135 

Division  of  Decimals, 136 

DECIMAL    CURRENCY. 

Notation  and  Numeration  of  Decimal  Currency, 139 

Reduction  of  Decimal  Currency, 141 

Addition  of  Decimal  Currency, 142 

Subtraction  of  Decimal  Currency, 143 

Multiplication  of  Decimal  Currency, 144 

Division  of  Decimal  Currency, 145 

Additional  Applications, 147 

When  the  Price  is  an  Aliquot  Part  of  a  Dollar, 147 

To  find  the  Cost  of  a  Quantity, 148 

To  find  the  Price  of  One, 149 

To  find  the  Quantity 149 

Articles  sold  by  the  100  or  1000,  150 

Articles  sold  by  the  Ton, 151 

Bills / 152 

Promiscuous  Examples 154 

COMPOUND    NUMBERS. 

Reduction, 158 

Definitions,  &c 7 158 

English  Money, 159 

Troy  Weight 161 


CONTENTS.  Vll 

Page 

Apothecaries'  Weight, 162 

Avoirdupois  Weight 163 

Long  Measure, 166 

Surveyors'  Long  Measure, 168 

Square  Measure,   169 

Surveyors'  Square  Measure, 172 

Cubic  Measure, 173 

Liquid  Measure, 175 

Dry  Measure, 176 

Time, 178 

Circular  Measure, 180 

Counting ;  Paper ;  Books,  &c., 181 

Reduction  of  Denominate  Fractions, 183 

Addition  of  Compound  Numbers, 190 

Addition  of  Denominate  Fractions, 193 

Subtraction 194 

To  find  the  Difference  in  Dates, 196 

Table, 197 

Subtraction  of  Denominate  Fractions, 198 

Multiplication  of  Compound  Numbers, 199 

Division, 201 

Longitude  and  Tune 203 

Duodecimals 206 

Promiscuous  Examples, 210 


PERCENTAGE. 

Definitions,  &c., 213 

Commission  and  Brokerage, 218 

Stocks 222 

Profit  and  Loss, 225 

Insurance, 231 

Taxes, 232 

Custom  House  Business, 235 

Simple  Interest, 239 

Partial  Payments  or  Indorsements, 245 

Problems  in  Interest, 251 


Vlll  CONTENTS. 

Page 

Compound  Interest 254 

Discount, 257 

Banking, 260 

Exchange, 264 

Equation  of  Payments, 269 

RATIO    AND    PROPORTION. 

Ratio 277 

Proportion, 280 

Simple  Proportion 281 

Compound  Proportion, 287 

Partnership, 292 

Analysis, 296 

Alligation  Medial, 305 

Alligation  Alternate, 306 

Involution, .  311 

Evolution, 312 

Square  Root, 313 

Cube  Root, 320 

Arithmetical  Progression, 326 

Geometrical  Progression, 329 

Promiscuous  Examples, 332 


PRACTICAL   ARITHMETIC. 


DEFINITIONS. 

1.  Quantity  is  any  thing  that  can  be  increased,  diminished, 
or  measured. 

2.  Mathematics  is  the  science  of  quantity. 

3.  A  Unit  is  one,  or  a  single  thing. 

4.  A  Number  is  a  unit,  or  a  collection  of  units. 
o.  An  Integer  is  a  whole  number. 

6.  The  Unit  of  a  Number  is  one  of  the  same   kind  or 
name  as  the  number.     Thus,  the  unit  of  23  is  1 ;  of  23  dollars, 
1  dollar ;  of  23  feet,  1  foot. 

7.  Like  Numbers  have  the  same  kind  of  unit.     Thus,  74, 
16,  and  250 ;  7  dollars  and  62  dollars ;  19  pounds,  320  pounds, 
and  86  pounds ;  4  feet  6  inches,  and  17  feet  9  inches. 

8.  An  Abstract  Number  is  a  number  used  without  refer- 
ence to  any  particular  thing  or  quantity.     Thus,  17 ;  365 ;  8540. 

9.  A  Concrete  Number  is  a  number  used  with  reference 
to  some  particular  thing  or  quantity.     Thus,  17  dollars ;   365 
days ;  8540  men. 

NOTES.    1.   The  unit  of  an  abstract  number  is  1,  and  is  called  Unity. 
2.    Concrete  numbers  are,   by   some,    called   Denominate  Numbers. 
Denomination  means  the  name  of  the  unit  of  a  concrete  number. 

10.  Arithmetic  is  the  Science  of  numbers,  and  the  Art  of 
computation. 

11.  A  Sign  is  a  character  indicating  an  operation  to  be 
performed. 

1£.    A  Rule  is  a  prescribed  method  of  performing  an  op- 
eration. 

Define  quantity.     Mathematics.     A  unit.     A  number.  An  integer. 

The  unit  of  a  number.     Like   numbers.     An   abstract  number.     A 

concrete   number.     The  unit  of  an  abstract  number.  Denominate 
numbers.     Arithmetic.     A  sign,  or  symbol.     A  rule. 


10  SIMPLE   NUMBERS. 


NOTATION  AND   NUMERATION. 

13.  Notation  is  a  method  of  writing  or  expressing  numbers 
by  characters ;  and, 

14:.  Numeration  is  a  method  of  reading  numbers  expressed 
by  characters. 

15.  Two  systems  of  notation  are  in  general  use  —  the 
Roman  and  the  Arabic. 

NOTE.  The  Roman  Notation  is  supposed  to  have  been  first  used 
by  the  Romans ;  hence  its  name.  The  Arabic  Notation  was  intro- 
duced into  Europe  by  the  Arabs,  by  whom  it  was  supposed  to  have 
been  invented.  But  investigations  have  shown  that  it  was  adopted  by 
them  about  600  years  ago,  and  that  it  has  been  in  use  among  the  Hin- 
doos more  than  2000  years.  From  this  latter  fact  it  is  sometimes 
called  the  Indian  Notation. 

THE   ROMAN   NOTATION 

1C.  Employs  seven  capital  letters  to  express  numbers, 
thus  : 

Letters,         I          V          X          L          C          D          M 


Values,          one,          five,  ten,  fifty,      h  °5*  d.  hnSj 


one 
hundred,  hundred,  thousand. 


17.  The  Roman  notation  is  founded  upon  five  principles, 
as  follows : 

1st.  Repeating  a  letter  repeats  its  value.  Thus,  II  repre- 
sents two,  XX  twenty,  CCC  three  hundred. 

2d.  If  a  letter  of  any  value  be  placed  after  one  of  greater 
value,  its  value  is  to  be  united  to  that  of  the  greater.  Thus, 
XI  represents  eleven,  LX  sixty,  DC  six  hundred. 

3d.  If  a  letter  of  any  value  be  placed  before  one  of  greater 
value,  its  value  is  to  be  taken  from  that  of  the  greater.  Thus, 
IX  represents  nine,  XL  forty,  CD  four  hundred. 

Define  notation.     Numeration.     What  systems  of  notation  are  now 
in  general  use  ?     From  what  are  their  names  derived  ?     What  ar> 
to  express  numbers  in  the  Roman  notation  ?     What  is  the  value  of  each  ? 
What  is  the  first  principle  of  combination  ?     Second  ?     Third  ? 


NOTATION   AND   NUMERATION.  11 

4th.  If  a  letter  of  any  value  be  placed  between  two  letters, 
each  of  greater  value,  its  value  is  to  be  taken  from  the  united 
value  of  the  other  two.  Thus,  XIV  represents  fourteen, 
XXIX  twenty-nine,  XCIV  ninety-four. 

5th.  A  bar  or  dash  placed  over  a  letter  increases  its  value 
one  thousand  fold.  Thus,  V  signifies  five,  and  V  five  thou- 
sand ;  L  fifty,  and  L  fifty  thousand. 

TABLE    OF   ROMAN    NOTATION. 

I  is  One.  XX  is  Twenty. 

II  "  Two.  XXI  "  Twenty-one. 

III  "  Three.  XXX  "   Thirty. 

IV  "  Four.  XL  "  »Forty. 
V  "  Five.  L  "   Fifty. 

VI  "  Six.  LX  "  Sixty. 

VII  "  Seven.  LXX  "  Seventy. 

VIII  "  Eight.  LXXX  "  Eighty. 

IX  "  Nine.  XC  «  Ninety. 

X  «  Ten.  C  "  One  hundred. 

XI  "  Eleven.  CC  "  Two  hundred. 

XII  "  Twelve.      .  D  "  Five  hundred. 

XIII  "  Thirteen.  DC  "  Six  hundred. 

XIV  "  Fourteen.  M  "  One  thousand,      [dred. 
XV  "  Fifteen.  MC  "  One  thousand  one  hun- 

XVI  "  Sixteen.  MM  "  Two  thousand. 

XVII  "  Seventeen.  ^X  "  Ten  thousand. 

XVIII  "  Eighteen.  ~C  "  One  hundred  thousand. 

XIX  "  Nineteen.  M  "  One  million. 

NOTE.  The  system  of  Roman  notation  is  not  well  adapted  to  the 
purposes  of  numerical  calculation ;  it  is  principally  confined  to  the 
numbering  of  chapters  and  sections  of  books,  public  documents,  &c. 

Express  the  following  numbers  by  letters : 

1.  Eleven.  Ans.     XL 

2.  Fifteen.  Ans. 

Fourth  ?     Fifth  ?     Repeat  the  table.     What  is  the  value  of  LVTI  ? 

CLxxm?       xcvin^_    CDXXXII?       xcix-       DCXIX? 

VMDCCXLIX  ?      MDXXVCDLXXXIX  ?     To    what    uses    is    the 
Roman  notation  now  principally  confined  ? 


12  SIMPLE    NUMBERS. 

3.  Twenty-five. 

4.  Thirty-nine. 

5.  Forty-eight. 

6.  Seventy-seven. 

7.  One  hundred  fifty-nine. 

8.  Five  hundred  ninety-four. 

9.  One  thousand  five  hundred  thirty-eight. 

10.  One  thousand  nine  hundred  ten. 

11.  Express  the  present  year. 

THE    ARABIC    NOTATION 

18.  Employs  ten  Characters  or  figures  to  express  numbers. 
Thus, 

Figures,  0123456789 

Nimcs  and  }    naught    one,     two,    three,  four,    five,     six,     seven,   eight,    nine. 

C         or 
values,          $  ciphcr> 

19.  The  first  character  is  called  naught,  because  it  has  no 
value  of  its  own.     The  other  nine  characters  are  called  signif- 
icant figures,  because  each  has  a  value  of  its  own. 

SO.  The  significant  figures  are  also  called  Digits,  a  word  de- 
rived from  the  Latin  term  digitus,  which  signifies  finger. 

21.    The  naught  or  cipher  is  also  called  nothing,  and  zero. 

The  ten  Arabic  characters  are  the  Alphabet  of  Arithmetic, 
and  by  combining  them  according  to  certain  principles,  all 
numbers  can  be  expressed.  We  will  now  examine  the  most 
important  of  these  principles.* 

%%.  Each  of  the  nine  digits  has  a  value  of  its  own  ;  hence 
any  number  not  greater  than  9  can  be  expressed  by  one 
figure. 

*  Fractional  and  decimal  notation,  and  the  notation  of  compound  numbers,  will  be 
discussed  in  their  appropriate  places. 

What  are  used  to  express  numbers  in  the  Arabic  notation  ?  What 
is  the  value  of  each  ?  What  general  name  is  given  to  the  significant 
figures  ?  Why  ?  Numbers  less  than  ten,  how  expressed  ? 


NOTATION   AND   NUMERATION.  13 

S3.  As  we  have  no  single  character  to  represent  ten,  we 
express  it  by  writing  the  unit,  1,  at  the  left  of  the  cipher,  0? 
thus,  10.  In  the  same  manner  we  represent 

2  tens,  3  tens,  4  tens,  5  tens,  6  tens,  7  tens,  8  tens,  9  tens, 

or              or  or          or  or  or              or               or 

twenty,  thirty,  forty,     fifty,     sixty,  seventy,  eighty,  ninety, 

20;          30;  40;      50;       60;  70;           80;         90. 

24.  When  a  number  is  expressed  by  two  figures,  the  right 
hand  figure  is  called  units,  and  the  left  hand  figure  tens. 

We  express  the  numbers  between  10  and  20  by  writing 
the  1  in  the  place  of  tens,  with  each  of  the  digits  respectively 
in  the  place  of  units.  Thus, 

eleven,   twelve,    thirteen,  fourteen,    fifteen,   sixteen,  seventeen,  eighteen,  nineteen. 

11,       12,       13,       14,        15,      16,        17,         18,        19. 

In  like  manner  we  express  the  numbers  between  20  and 
30,  between  30  and  40,  and  between  any  two  successive  tens. 
Thus,  21,  22,  23,  24,  25,  26,  27,  28,  29,  34,  47,  56,  72,  93. 
The  greatest  number  that  can  be  expressed  by  two  figures 
is  99. 

SJt5.  We  express  one  hundred  by  writing  the  unit,  1,  at  the 
left  hand  of  two  ciphers,  or  the  number  10  at  the  left  hand  of 
one  cipher;  thus,  100.  In  like  manner  we  write  two  hun- 
dred, three  hundred,  &c.,  to  nine  hundred.  Thus, 

one       two       three      four        five         six       seven     eight      nine 
hundred,  hundred,  hundred,  hundred,  hundred,  hundred,  hundred,  hundred,  hundred, 

100,     200,     300,     400,     500,      600,     700,     800,     900. 

26.  When  a  number  is  expressed  by  three  figures,  the 
right  hand  figure  is  called  units,  the  second  figure  tens,  and 
the  left  hand  figure  hundreds. 

As  the  ciphers  have,  of  themselves,  no  value,  but  are  always 
used  to  denote  the  absence  of  value  in  the  places  they  occupy, 

Tens,  how  expressed  ?  The  right  hand  figure  called  what  ?  Left 
hand  figure,  what  ?  What  is  the  greatest  number  that  can  be  expressed 
by  two  figures  ?  One  hundred,  how  expressed  ?  When  numbers  are 
expressed  by  three  figures,  what  names  are  given  to  each  5 


14  SIMPLE    NUMBERS. 

we  express  tens  and  units  with  hundreds,  by  writing,  in  place 
of  the  ciphers,  the  numbers  representing  the  tens  and  units. 
To  express  one  hundred  fifty  we  write  1  hundred,  5  tens,  and 
0  units;  thus,  150.  To  express  seven  hundred  ninety-two, 
we  write  7  hundreds,  9  tens,  and  2  units  ;  thus, 


I  I  1 

792 

The  greatest  number  that  can  be  expressed  by  three  figures 
is  999. 

EXAMPLES   FOR   PRACTICE. 

1.  Write  one  hundred  twenty-five. 

2.  Write  four  hundred  eighty-three. 

3.  Write  seven  hundred  sixteen. 

4.  Express  by  figures  nine  hundred. 

5.  Express  by  figures  two  hundred  ninety. 

6.  Write  eight  hundred  nine. 

7.  Write  five  hundred  five. 

8.  Write  five  hundred  fifty-seven. 

ST.  We  express  one  thousand  by  writing  the  unit,  1,  at 
the  left  hand  of  three  ciphers,  the  number  10  at  the  left  hand 
of  two  ciphers,  or  the  number  100  at  the  left  hand  of  one 
cipher ;  thus,  1000.  In  the  same  manner  we  write  two 
thousand,  three  thousand,  &c.,  to  nine  thousand  ;  thus, 

one        two       three      four       five        six        seven     eight      nine 

thousand,  thousand,  thousand,  thousand,  thousand,  thousand,  thousand,  thousand,  thousand. 

1000,  2000,  3000,  4000,  5000,  6000,  7000,  8000,  9000. 

S8.  When  a  number  is  expressed  by  four  figures,  the 
places,  commencing  at  the  right  hand,  are  units,  tens,  hundreds, 
thousands. 

Use  of  the  cipher,  what  ?  Greatest  number  that  can  be  expressed  by 
three  figures  ?  One  thousand,  how  expressed  ?  How  many  figures 
used  ?  Names  of  each  ? 


NOTATION  AND  NUMERATION.  15 

To  express  hundreds,  tens,  and  units  with  thousands,  we 
write  in  each  place  the  figure  indicating  the  number  we  wish 
to  express  in  that  place.  To  write  four  thousand  two  hun- 
dred sixty-nine,  we  write  4  in  the  place  of  thousands,  2  in  the 
place  of  hundreds,  6  in  the  place  of  tens,  and  9  in  the  place  of 
units;  thus, 


I  I  1  I 


4t      z       o       v 

The  greatest  number  that  can  be  expressed  by/owr  figures 
f)999. 


EXAMPLES    FOR   PRACTICE. 

Express  the  following  numbers  by  figures  :  — 

1.  One  thousand  two  hundred. 

2.  Five  thousand  one  hundred  sixty. 

3.  Three  thousand  seven  hundred  forty-one. 

4.  Eight  thousand  fifty-six. 

5.  Two  thousand  ninety. 

6.  Seven  thousand  nine. 

7.  One  thousand  one. 

8.  Nine  thousand  four  hundred  twenty-seven. 

9.  Four  thousand  thirty-five. 

10.  One  thousand  nine  hundred  four. 
Head  the  following  numbers  :  — 

11.  76;      128;      405;      910;      116;    3416;    1025. 

12.  2100;  5047;  7009;  4670;  3997;  1001. 


Next  to  thousands  come  tens  of  thousands,  and  next 
to  these  come  hundreds  of  thousands,  as  tens  and  hundreds 
come  in  their  order  after  units.  Ten  thousand  is  expressed 
by  removing  the  unit,  1,  one  place  to  the  left  of  the  place 

Greatest  number  expressed  by  four  figures  r     Tens  of  thousands,  how 
expressed  ?     Hundreds  of  thousands  ? 


16  SIMPLE   NUMBERS. 

of  thousands,  or  by  writing  it  at  the  left  hand  of  four  ci- 
phers; thus,  10000;  and  one  hundred  thousand  is  expressed 
by  removing  the  unit,  1,  still  one  place  further  to  the  left,  or 
by  writing  it  at  the  left  hand  of  five  ciphers ;  thus,  100000. 
"We  can  express  thousands,  tens  of  thousands,  and  hundreds  of 
thousands  in  one  number,  in  the  same  manner  as  we  express 
units,  tens,  and  hundreds  in  one  number.  To  express  five 
hundred  twenty-one  thousand  eight  hundred  three,  we  write  5  in 
the  sixth  place,  counting  from  units,  2  in  the  fifth  place,  1  in 
the  fourth  place,  8  in  the  third  place,  0  in  the  second  place, 
(because  there  are  no  tens,)  and  3  in  the  place  of  units; 
thus, 


II  SJ  I  I  1  1 

5         21803 

The  greatest  number  that  can  be  expressed  by  five  figures 
is  99999  ;  and  by  six  figures,  999999. 

EXAMPLES    FOB   PRACTICE. 

Write  the  following  numbers  in  figures :  — 

1.  Twenty  thousand. 

2.  Forty-seven  thousand. 

3.  Eighteen  thousand  one  hundred. 

4.  Twelve  thousand  three  hundred  fifty. 

5.  Thirty-nine  thousand  five  hundred  twenty-two. 

6.  Fifteen  thousand  two  hundred  six. 

7.  Eleven  thousand  twenty-four. 

8.  Forty  thousand  ten. 

9.  Sixty  thousand  six  hundred. 

10.  Two  hundred  twenty  thousand. 

11.  One  hundred  fifty-six  thousand. 

12.  Eight  hundred  forty  thousand  three  hundred. 

Greatest  number  expressed  by  five  figures  ?     Six  figures  ? 


NOTATION  AND  NUMERATION.  17 

13.  Five  hundred  one  thousand  nine  hundred  sixty -four. 

14.  One  hundred  thousand  one  hundred. 

15.  Three  hundred  thirteen  thousand  three  hundred  thir- 
teen. 

16.  Seven  hundred  eighteen  thousand  four. 

17.  One  hundred  thousand  ten. 
Read  the  following  numbers :  — 

18.  5006;        12304;     96071;          5470;     203410. 

19.  36741;      400560;     13061;        49000;     100010. 

20.  200200;        75620;     90402;      218094;     100101. 

For  convenience  in  reading  large  numbers,  we  may  point 
them  off,  by  commas,  into  periods  of  three  figures  each,  count- 
ing from  the  right  hand  or  unit  figure.  This  pointing  enables 
us  to  read  the  hundreds,  tens,  and  units  in  each  period  with 
facility. 

3O.  Next  above  hundreds  of  thousands  we  have,  succes- 
sively, units,  tens,  and  hundreds  of  millions,  and  then  follow 
units,  tens,  and  hundreds  of  each  higher  name,  as  seen  in  the 
following 

NUMERATION   TABLE. 

I    |     |      j       .      f      .     | 

CM  CM  ^*-(  CM  tt_i  tj-j  ft  i  ti_.  Ci_i 

000  000000 

r~^ 

4 


98,765,432,109,876,556,789,012,345 

ninth  eighth  seventli   sixth      fifth      fourth    third    second     first 
period,  period,  period,  period,  period,  period,  period,  period,  period. 

How  may  figures  be  pointed  off?  One  million,  how  expressed? 
Next  period  above  millions,  what  ?  Give  the  name  of  each  successive 
period. 

B* 


18  SIMPLE    NUMBERS. 

NOTE.  This  is  called  the  French  method  of  pointing  off  the  peri- 
ods, and  is  the  one  in  general  use  in  this  country. 

31.  Figures  occupying  different  places  in  a  number,  as 
units,  tens,  hundreds,  &c.,  are  said  to  express  different  orders 
of  units. 

Simple  units          are  called  units  of  the  first     order. 
Tens  "       "          "      "    "    second     " 

Hundreds  «       "          «      "    "    third       " 

Thousands  «       «          "      "    «  fourth     « 

Tens  of  thousands  "       "          "      "    "  fifth        " 

and  so  on.  Thus,  452  contains  4  units  of  the  third  order,  5 
units  of  the  second  order,  and  2  units  of  the  first  order. 
1,030,600  contains  1  unit  of  the  seventh  order,  (millions,)  3 
units  of  the  fifth  order,  (tens  of  thousands,)  and  6  units  of  the 
third  order,  (hundreds.) 

EXAMPLES    FOR    PRACTICE. 

Write  and  read  the  following  numbers  :  — 

1.  One  unit  of  the  third  order,  four  of  the  second. 

2.  Three  units  of  the  fifth  order,  two  of  the  third,  one  of  the 
first. 

3.  Eight  units  of  the  fourth  order,  five  of  the  second. 

4.  Two  units  of  the  seventh  order,  nine  of  the  sixth,  four 
of  the  third,  one  of  the  second,  seven  of  the  first. 

5.  Three  units  of  the  sixth  order,  four  of  the  second. 

6.  Nine  units  of  the  eighth  order,  six  of  the  seventh,  three 
of  the  fifth,  seven  of  the  fourth,  nine  of  the  first. 

7.  Four  units  of  the  tenth  order,  six  of  the  eighth,  four  of 
the  seventh,  two  of  the  sixth,  one  of  the  third,  five  of  the  sec- 
ond. 

8.  Eight  units  of  the  twelfth  order,  four  of  the  eleventh,  six 
of  the  tenth,  nine  of  the  seventh,  three  of  the  sixth,  five  of  the 
fifth,  two  of  the  third,  eight  of  the  first. 


Units  of  different  orders  are  what  ? 


NOTATION   AND    NUMERATION.  19 

• 

32.  From  the  foregoing  explanations  and  illustrations,  we 
derive  several  important  principles,  which  we  will  now  pre- 
sent. 

1st.  Figures  have  two  values,  Simple  and  Local. 

The  Simple  Value  of  a  figure  is  its  value  when  taken  alone ; 
thus,  2,  5,  8. 

The  Local  Value  of  a  figure  is  its  value  when  used  with  an- 
other figure  or  figures  in  the  same  number ;  thus,  in  842  the 
simple  values  of  the  several  figures  are  8,  4,  and  2 ;  but  the 
local  value  of  the  8  is  800 ;  of  the  4  is  4  tens,  or  40 ;  and  of 
the  2  is  2  units. 

NOTE.  When  a  figure  occupies  units'  place,  its  simple  and  local 
values  are  the  same. 

2d.  A  digit  or  figure,  if  used  in  the  second  place,  expresses 
tens ;  in  the  third  place,  hundreds ;  in  the  fourth  place,  thou- 
sands ;  and  so  on. 

3d.  As  10  units  make  1  ten,  10  tens  1  hundred,  10  hun- 
dreds 1  thousand,  and  10  units  of  any  order,  or  in  any  place, 
make  one  unit  of  the  next  higher  order,  or  in  the  next  place 
at  the  left,  we  readily  see  that  the  Arabic  method  of  notation 
is  based  upon  the  following 

TWO    GENERAL    LAWS. 

I.  Numbers  increase  from  right  to  left,  and  decrease  from 
left  to  right,  in  a  tenfold  ratio. 

II.  Every  removal  of  a  figure  or  number  one  place  to  the 
left,  increases  its  local  value  tenfold  ;  and  every  removal  of  a 
figure  or  number  one  place  to  the  right,  diminishes  its  local 
value  tenfold. 

Thus, 

6  is     6  units. 

60  is  10  times  6  units. 

600  is  10  times  6  tens. 

6000  is  10  times  6  hundreds. 

60000  is  10  times  6  thousands. 

First  principle  derived  ?  What  is  the  simple  value  of  a  figure  ?  Local  ? 
Second  principle  ?  Third  ?  First  law  of  Arabic  notation  ?  Second  ? 


20  SIMPLE   NUMBERS. 

4th.  The  local  value  of  a  figure  depends  upon  its  place  from 
units  of  the  first  order,  not  upon  the  value  of  the  figures  at  the 
right  of  it.  Thus,  in  425  and  400,  the  value  of  the  4  is  the 
same  in  both  numbers,  being  4  units  of  the  third  order,  or  4 
hundred. 

NOTE.  Care  should  be  taken  not  to  mistake  the  local  value  of  a 
figure  for  the  value  of  the  whole  number.  For,  although  the  value  of 
the  4  (hundreds)  is  the  same  in  the  two  numbers,  425  and  400,  the  value 
of  the  whole  of  the  first  number  is  greater  than  that  of  the  second. 

5th.  Every  period  contains  three  figures,  (units,  tens,  and 
hundreds,)  except  the  left  hand  period,  which  sometimes  con- 
tains only  one  or  two  figures,  (units,  or  units  and  tens.) 

33.  As  we  have  now  analyzed  all   the  principles  upon 
which  the  writing  and  reading  of  whole  numbers  depend,  we 
will  present  these  principles  in  the  form  of  rules. 

RULE    FOR    NOTATION. 

L  Beginning  at  the  left  hand,  write  the  figures  belonging  to 
the  highest  period. 

II.  Write  the  hundreds,  tens,  and  units  of  each  successive 
period  in  their  order,  placing  a  cipher  wherever  an  order  of 
units  is  omitted. 

RULE    FOR   NUMERATION. 

L  Separate  the  number  into  periods  of  three  figures  each, 
commencing  at  the  right  hand. 

II.  Beginning  at  the  left  hand,  read  each  period  separately, 
and  give  the  name  to  each  period,  except  the  last,  or  period 
of  units. 

34.  Until  the  pupil  can  write  numbers  readily,  it  may  be 
well  for  him  to  write  several  periods  of  ciphers,  point  them  off, 
over  each  period  write  its  name,  thus, 

Trillions,    Billions,    Millions,    Thousands,    Units. 

000  ,    000,    000,     000  ,    000 

Fourth  principle  ?  What  caution  is  given  ?  Fifth  principle  ?  Rule 
for  notation  ?  Numeration  ? 


NOTATION   AND   NUMERATION 

and  then  write  the  given  numbers  underneWBf  ifftlef?  J^fe-S  , 
priate  places.  Vv^d  «.    °-BI 

x^vsf  r  »—  —  »*  •»  p 

EXERCISES    IN   NOTATION   AND    NUMERATION. 

Express  the  following  numbers  by  figures  :  — 

1.  Four  hundred  thirty-six. 

2.  Seven  thousand  one  hundred  sixty-four. 

3.  Twenty-six  thousand  twenty-six. 

4.  Fourteen  thousand  two  hundred  eighty. 

5.  One  hundred  seventy-six  thousand. 

6.  Four  hundred  fifty  thousand  thirty-nine. 

7.  Ninety-five  million. 

8.  Four  hundred  thirty-three  million  eight  hundred  sixteen 
thousand  one  hundred  forty-nine. 

9.  Nine  hundred  thousand  ninety. 

10.  Ten  million  ten  thousand  ten  hundred  ten. 

11.  Sixty-one  billion  five  million. 

12.  Five  trillion  eighty  billion  nine  million  one. 

Point  off,  numerate,  and  read  the  following  numbers :  — 


13.   8240. 
14.  400900. 
15.    308. 
16.  60720. 

17.      1010. 
18.  57468139. 
19.      5628. 
20.  850026800. 

21.      370005. 
22.   9400706342. 
23.    38429526. 
24.  74268113759. 

25.  Write  seven  million  thirty-six. 

26.  Write  five  hundred  sixty-three  thousand  four. 

27.  Write  one  million  ninety-six  thousand. 

28.  Numerate  and  read  9004082501. 

29.  Numerate  and  read  2584503962047. 

30.  A  certain  number  contains  3  units  of  the  seventh  order, 
6  of  the  fifth,  4  of  the  fourth,  1  of  the  third,  5  of  the  second, 
and  2  of  the  first ;  what  is  the  number  ? 

31 .  What  orders  of  units  are  contained  in  the  number  290648  ? 

32.  What   orders   of  units  are  contained   in  the  number 
1037050? 


22  SIMPLE   NUMBERS. 


ADDITION. 

MENTAL    EXERCISES. 

35.    1.    Henry  gave  5  dollars  for  a  vest,  and  7  dollars  for 
a  coat ;  how  much  did  he  pay  for  both  ? 

ANALYSIS.     He  gave  as  many  dollars  as  5  dollars  and  7  dollars, 
which  are  12  dollars.     Therefore  he  paid  12  dollars  for  both. 

2.  A  farmer  sold  a  pig  for  3  dollars,  and  a  calf  for  8  dol- 
lars ;  how  much  did  he  receive  for  both  ? 

3.  A  drover  bought  5  sheep  of  one  man,  9  of  another,  and 
3  of  another  ;  how  many  did  he  buy  in  all  ? 

4.  How  many  are  2  and  6?  2  and  7  ?   2  and  9  ?  2  and  8  ? 
2  and  10  ? 

5.  How  many  are  4  and  5  ?  4  and  8  ?  4  and  7  ?  4  and  9  ? 
.6.    How  many  are  6  and  4?   6  and  6  ?   6  and  9  ?   6  and  7  ? 

7.  How  many  are  7  and  7  ?  7  and  6  ?  7  and  8  ?  7  and  10  ? 
7  and  9  ? 

8.  How  many  are  5  and  4  and  6  ?  7  and  3  and  8  ?  6  and  9 
and  5? 

3G.    From  the  preceding  operations  we  perceive  that 
Addition  is  the  process  of  uniting  several  numbers  of  the 
same  kind  into  one  equivalent  number. 

37.  The  Sum  or  Amount  is  the  result  obtained  by  the 
process  of  addition. 

38.  The  sign,  -(-,  is  called  plus,  which  signifies  more. 
When  placed  between  two  numbers,  it  denotes  that  they  are 
to  be  added ;  thus,  6  -\-  4,  shows  that  6  and  4  are  to  be  added. 

39.  The  sign,  =,  is  called  the  sign  of  equality.     When 
placed  between  two  numbers,  or  sets  of  numbers,  it  signifies 
that   they   are    equal   to    each   other;    thus,   the    expression 
6  +  4  =  10,  is  read  6  phis  4  is  equal  to  10,  and  denotes  that 
the  numbers  6  and  4,  taken  together,  equal  the  number  10. 

Define  addition.     The   sum   or   amount  ?     Sign   of  addition  ?     Of 
equality  ? 


ADDITION. 


23 


ADDITION    TABLE. 


1- 

-    1=   2 

2+    1=   3 

3+    1=   4 

4+    1=   5 

1- 

-   2=   3 

2+   2=   4 

3+    2=   5 

4+   2=   6 

1- 

-   3=   4 

2- 

-    3  =:    5 

3+   3=   6 

4+   3=   7 

1- 

-   4=   5 

2- 

-   4=   6 

3+   4=    7 

4+   4=   8 

1- 

-   5=   6 

2- 

-   5=    7 

3+    5=   8 

4- 

-   5=    9 

1- 

-   6=   7 

2- 

-   6=    8 

3+    6=   9 

4J 

-6  =  10 

1- 

-   7=   8 

2- 

-   7=   9 

3+    7  =  10 

4- 

-   7  =  11 

1- 

-   8=   9 

2- 

-8  =  10 

3- 

-    8  =  11 

4- 

-   8  =  12 

1+   9  =  10 

2+   9  =  11 

3- 

-9  =  12 

4- 

-   9  =  13 

1  +  10  =  11 

2  +  10  =  12 

3- 

-10  =  13 

4- 

-10  =  14 

1  +  11  =  12 

2  +  11  1=  13 

3- 

-11  =  14 

4- 

-11  =  15 

1  +  12  =  13 

2  +  12  =  14 

3- 

-12  =  15 

4  +  12  =  16 

5- 

-   1=   6 

6+    1=   7 

7+    1=   8 

8+   1=   9 

5- 

-   2=   7 

6+   2  =  -8 

7+    2=   9 

8+   2  =  10 

5- 

-   3=   8 

6+   3=   9 

7+    3  =  10 

8+   3  =  11 

5- 

-   4=   9 

6+   4  =  10 

7+   4  =  11 

8+   4=12 

5+   5  =  10 

6+   5  =  11 

7+    5  =  12 

8+   5  =  13 

5- 

-   6  =  11 

6+   6  =  12 

7+   6  =  13 

8+   6  =  14 

5- 

-7  =  12 

6+    7  =  13 

7+    7  =  14 

8+   7  =  15 

5- 

-8  =  13 

6- 

-8  =  14 

7+   8  =  15 

8+   8  =  16 

5- 

-9  =  14 

6- 

-9  =  15 

7+    9  =  16 

8+   9  =  17 

5- 

-10  =  15 

6- 

-10  =  16 

7  +  10  =  17 

8  +  10  =  18 

5  +  11  =  16 

6- 

-11  =  17 

7  +  11  =  18 

8  +  11  =  19 

5  +  12  =  17 

6- 

-12  =  18 

7  +  12  =  19 

8  +  12  =  20 

9+    1  =  10 

10+    1  =  11 

11- 

-1  =  12 

12- 

1-    1  =  13 

9+    2  =  11 

10+   2  =  12 

11- 

-   2  =  13 

12- 

-2  =  14 

9+   3  =  12 

10- 

-3  =  13 

11- 

-3  =  14 

12- 

-3  =  15 

9- 

-4  =  13 

10- 

-4  =  14 

11- 

-   4  =  15 

12- 

-   4  =  16 

9- 

-5  =  14 

10- 

-5  =  15 

11- 

-5  =  16 

12- 

-   5  =  17 

9- 

-   6  =  15 

10+   6  =  16 

11+   6  =  17 

12- 

-   6  =  18 

9- 

-7  =  16 

10+   7  =  17 

11- 

-7  =  18 

12+   7  =  19 

9- 

-8  =  17 

10+   8  =  18 

11- 

-8  =  19 

12+   8  =  20 

9- 

-9  =  18 

10- 

-9=19 

11- 

-   9  =  20 

12+   9  =  21 

9- 

-10  =  19 

10- 

-  10  =  20 

11- 

-10  =  21 

12  +  10  =  22 

9  +  11  =  20 

10- 

-11=21 

11- 

-11=22 

12  +  11=23 

9  +  12  =  21 

10- 

-12  =  22 

11- 

-12  =  23 

12  +  12  =  24 

24  SIMPLE  NUMBERS. 

CASE    I. 

40.  When  the   amount  of   each  column  is  less 
than  10. 

1.  A  farmer  sold  some  hay  for  102  dollars,  six  cows  for 
162  dollars,  and  a  horse  for  125  dollars ;  how  much  did  he  re- 
ceive for  all  ? 

OPEKATION.         ANALYSIS.    We  arrange  the  numbers  so 

,3^  that  units  of  like  order  shall  stand  in  the 

same   column.     We  then  add  the  columns 

102  separately,  for  convenience   commencing  at 

162  the  right  hand,  and  write  each  result  under 

J25  the  column  added.     Thus,  we  have  5  and  2 

and  2  are  9,  the  sum  of  the  units ;  2  and  6 

Amount,     389  are  8,  the  sum  of  the  tens  ;  1  and  1  and  1 

are  3,  the  sum  of  the  hundreds.     Hence,  the 

entire  amount  is  3  hundreds  8  tens  and  9  units,  or  389,  the  Answer. 

EXAMPLES    FOR   PRACTICE. 

(2.)  (3.)  (4.)  (5.) 

pounds.  rods.  cents.  days. 

132  245  312  437 

243  321  243  140 

324  132  412  321 
Ans.     699 

6.  What  is  the  sum  of  144,  321,  and  232  ?          Ans.   697. 

7.  What  is  the  amount  of  122,  333,  and  401  ?     Ans.   856. 

8.  What  is  the  sum  of  42,  103,  321,  and  32  ?     Ans.   498. 

9.  A  drover  bought  three  droves  of  sheep.     The  first  con- 
tained 230,  the  second  425,  and  the  third  340 ;  how  many 
sheep  did  he  buy  in  all?  Ans.   995. 

CASE   II. 

41.  When  the  amount  of  any  column  equals  or 
exceeds  10. 

1.   A  merchant  pays  725  dollars  a  year  for  the  rent  of  a 
Case  I  is  what  ?     Give  explanation.     Case  n  is  what  ? 


ADDITION.  25 

store,  475  dollars  for  a  clerk,  and  367  dollars  for  other  ex- 
penses ;  what  is  the  amount  of  his  expenses  ? 

OPERATION.  ANALYSIS.    Arranging  the  num- 

-3  ^  £  bers  as  in  Case  I,  we  first  add  the 

|||  column  of  units,  and  find  the  sum 

725  to  be  17  units,  which  is  1  ten  and 

475  7  units.     We  write  the  7  units  in 

gg7  the  place  of  units,  and  the  1  ten  in 

-  the  place  of  tens.     The  sum  of  the 

Sum  of  the  units,                   17  figures  in  the  column  of  tens  is  15 

Sum  of  the  tens,                  15  tens,  which  is   1   hundred,  and  5 

Sum  of  the  hundreds,         14  tens.     We  write  the  5  tens  in  the 

l 


Total  amount,  1567 

the  place  of  hundreds.     We  next 

add  the  column  of  hundreds,  and  find  the  sum  to  be  14  hundreds, 
which  is  1  thousand  and  4  hundreds.  We  write  the  4  hundreds  in 
the  place  of  hundreds,  and  1  thousand  in  the  place  of  thousands. 
Lastly,  by  uniting  the  sum  of  the  units  with  the  sums  of  the  tens 
and  hundreds,  we  find  the  total  amount  to  be  1  thousand  5  hundreds 
6  tens  and  7  units,  or  1567. 

This  example  may  be  performed  by  another  method,  which 
is  the  common  one  in  practice.  Thus  : 

OPERATION.        ANALYSIS.    Arranging  the  numbers  as  before,  we 

725          add  the  first  column  and  find  the  sum  to  be  17  units  ; 

475          writing  the  7  units  under  the  column  of  units,  we   add 

ggy          the  1  ten  to  the  column  of  tens,  and  find  the  sum  to  be 

16  tens  ;  writing  the  6  tens  under  the  column  tens,  we 

1567          add  the  1  hundred  to  the  column  of  hundreds,  and  find 

the  sum  to  be  15  hundreds  ;  as  this  is  the  last  column, 

we  write  down  its  amount,  15  ;  and  we  have  the  icJwle  amount,  1567, 

as  before. 

NOTES.  1.  Units  of  the  same  order  are  written  in  the  same  column  ; 
and  when  the  sum  in  any  column  is  10  or  more  than  10,  it  produces 
one  or  more  units  of  a  higher  order,  which  must  be  added  to  the  next 
column.  This  process  is  sometimes  called  "  carrying  the  tens." 

2.  In  adding,  learn  to  pronounce  the  partial  results  witHout  naming 
the  numbers  separately  ;  thus,  instead  of  saying  7  and  5  are  12,  and 
5  are  17,  simply  pronounce  the  results,  7,  12,  17,  &c. 

Give  explanation.  Second  explanation.  What  is  meant  by  carry- 
ing the  tens  ? 


26  SIMPLE   NUMBERS. 


.    From   the  preceding  examples  and  illustrations  we 
deduce  the  following 

RULE.  I.  Write  the  numbers  to  be  added  so  that  all  the  units 
of  the  same  order  shall  stand  in  the  same  column  ;  that  is,  units 
under  units,  tens  under  tens,  fyc. 

II.  Commencing  at  units,  add  each  column  separately,  and 
write  the  sum  underneath,  if  it  be  less  than  ten. 

III.  If  the  sum  of  any  column  be  ten  or  more  than  ten,  write 
the  unit  figure  only,  and  add  the  ten  or  tens  to  the  next  column. 

IV.  Write  the  entire  sum  of  the  last  column. 

PROOF.  1st.  Begin  with  the  right  hand  or  unit  column,  and 
add  the  figures  in  each  column  in  an  opposite  direction  from 
that  in  which  they  were  first  added  ;  if  the  two  results  agree, 
the  work  is  supposed  to  be  right.  Or, 

2d.  Separate  the  numbers  added  into  two  sets,  by  a  hori- 
zontal line  ;  find  the  sum  of  each  set  separately  ;  add  these 
sums,  and  if  the  amount  be  the  same  as  that  first  obtained,  the 
work  is  presumed  to  be  correct. 

NOTE.  By  the  methods  of  proof  here  given,  the  numbers  are  united 
in  new  combinations,  which  render  it  almost  impossible  for  two  pre- 
cisely similar  mistakes  to  occur. 

The  first  method  is  the  one  commonly  used  in  business. 

EXAMPLES   FOR   PRACTICE. 


(2.) 

miles. 

(3.) 

inches. 

(4.) 

tons. 

(5.) 

feet. 

(6.) 

bushels. 

24 

321 

427 

1342 

3420 

48 

479 

321 

7306 

7021 

96 

165 

903 

5254 

327 

82 

327 

278 

8629 

97 

250 

1292 

1929 

22531 

10865 

-    Rule,  first  step  ?    Second  ?    Third  ?     Fourth  ?    Proof,  first  method  ? 
Second  ?    Upon  what  principle  are  these  methods  of  proof  founded  ? 


ADDITION.  27 

(7.)  (8.)  (9.)  (10.) 

hours.  years.  gallons.  rods. 

347  7104  3462  47637 

506  3762  863  3418 

218  9325  479  703 

312  4316  84  26471 

424  2739  57  84 

11.  42  +  64  +  98  +  70  +  37  =  how  many  ?  Am.  311. 

12.  312  +  425  +  107  +  391  +  76  =  how  many  ? 

Am.  1311. 

13.  1476  +  375  -f  891  +  66  +  80  =  how  many  ? 

Ans.  2888. 

14.  37042  +  1379  +  809  +  127  +  40  =  how  many  ? 

Ans.  39397. 

15.  What  is  the  sum  of  one  thousand  six  hundred  fifty-six, 
eight  hundred  nine,  three  hundred  ten,  and  ninety-four  ? 

Ans.   2869. 

16.  Add  forty-two  thousand  two  hundred  twenty,  ten  thou- 
sand  one  hundred  five,  four  thousand  seventy-five,  and  five 
hundred  seven.  Ans.   56907. 

17.  Add  two  hundred  ten  thousand  four  hundred,  one  hun- 
dred thousand  five  hundred  ten,  ninety  thousand  six  hundred 
eleven,   forty-two   hundred    twenty-five,  and    eight   hundred 
ten.  Ans.   406556. 

18.  What  is  the  sum  of  the  following  numbers :  seventy- 
five,  one  thousand  ninety-five,  six  thousand  four  hundred  thir- 
ty-five, two  hundred  sixty-seven  thousand,  one  thousand  four 
hundred  fifty-five,  twenty-seven  million  eighteen,  two  hundred 
seventy  million  twenty-seven  thousand  ?    Ans.   297303078. 

19.  A  man  on  a  journey  traveled  the  first  day  37  miles, 
the  second  33  miles,  the  third  40  miles,  and  the  fourth  35  miles ; 
how  far  did  he  travel  in  the  four  days  ? 

20.  A  wine  merchant  has  in  one  cask  75  gallons,  in  another 
65,  in  a  third  57,  in  a  fourth  83,  in  a  fifth  74,  and  in  a  sixth 
67  ;  how  many  gallons  has  he  in  all?  Ans.   421. 


28  SIMPLE   NUMBERS. 

21.  An  estate  is  to  be  shared  equally  by  four  heirs,  and 
the  portion  to  each  heir  is  to  be  3754  dollars ;  what  is  the 
amount  of  the  estate?  Ans.    15016  dollars. 

22.  How  many  men  in  an  army  consisting  of  52714  in- 
fantry, 5110  cavalry,   6250  dragoons,  3927  light-horse,  928 
artillery,  250  sappers,  and  406  miners  ? 

23.  A  merchant  deposited  56  dollars  in  a  bank  on  Monday, 
74  on  Tuesday,  120  on  Wednesday,  96  on  Thursday,  170  on 
Friday,  and  50  on  Saturday ;  how  much  did  he  deposit  during 
the  week  ? 

24.  A  merchant  bought  at  public  sale  746  yards  of  broad- 
cloth, 650  yards  of  muslin,   2100  yards  of  flannel,  and  250 
yards  of  silk  ;  how  many  yards  in  all  ? 

25.  Five  persons  deposited  money  in  the  same  bank ;  the 
first,  5897  dollars;    the   second,  12980  dollars;    the   third, 
65973  dollars ;  the  fourth,  37345   dollars ;    and  the  fifth  as 
much  as  the  first  and  second  together ;  how  many  dollars  did 
they  all  deposit  ?  Ans.   141072  dollars. 

26.  A  man  willed  his  estate  to  his  wife,  two  sons,  and  four 
daughters ;  to  his  daughters  he  gave  2630  dollars  apiece,  to 
his  sons,  each  4647  dollars,  and  to  his  wife  3595  dollars; 
how  much  was  his  estate  ?  Ans.   23409  dollars. 

(27.)  (28.)  (29.)  (30.)  (31.) 

476  908  126  443  180 

390  371  324  298  976 

915  569  503  876  209 

207  245  891  569  314 

841  703  736  137  563 

632  421  517  910  842 

234  127  143  347  175 

143  354  274  256  224 

536  781  531  324  135 

245  436  275  463  253 


ADDITION. 


29 


(32.) 
3120 
8417 
2645 
9016 
1857 
4532 


(33.) 
14903 
32087 
63124 
51037 
86710 
39425 


(34.) 
40371 
56108 
92436 
15682 
23761 
58619 


(35.) 

43916 

16782 

58468 

20375 

3741 

897 


(36.) 
2981043 
7126459 
9412767 
7891234 
109126 
84172 
72120 
27676921 


(37.) 

1278976 

7654301 

876120 

723456 

31309 

4871 

978 


(38.) 

416785413 

915123460 

31810213 

7367985 

654321 

37853 

2685 

1371781930 


39.  Add  8635,  2190,  7421,  5063,  2196,  and  1245. 

Ans.  26750. 

40.  Add  246034,  29865,  47321,  58653,  64210,  5376,  9821, 
and  340.  Ans.  461620. 

41.  Add  27104,  32540,  10758,  6256,  704321,  730491, 
2787316,  and  2749104. 

42.  Add  1,  37,  29504,  6790312,  18757421,  and  265. 

Ans.  25577540. 

43.  Add  56263,  211964,  56321,  18536,  4340, 279,  and  73. 

Ans.  347776. 

44.  Add  3742  bushels,  493  bushels,  927  bushels,  643  bush- 
els, and  953  bushels. 

45.  Add  7346  acres,  9387  acres,  8756  acres,  8394  acres, 
and  32724  acres.  Ans.    66607. 

46.  Henry  received  at  one  time  15  apples,  at  another  115, 
and  at  another  19  ;  how  many  did  he  receive  ? 

c* 


30  SIMPLE   NUMBERS. 

47.  A  man  commenced  farming  at  the  west,  and  raised,  the 
first  year,  724  bushels  of  corn ;  the  second  year,  3498  bushels  ; 
the  third  year,  9872  bushels;  the  fourth  year,  9964  bushels; 
the  fifth  year,  11078  bushels;  how  many  bushels  did  he  raise 
in  the  five  years  ?  Ans.   35136  bushels. 

48.  A  has  3648  dollars,   B  has  7035  dollars,  C  has  429 
dollars  more  than  A  and  B  together,  and  D  has  as  many  dol- 
lars as  all  the  rest ;  how  many  dollars  has  D  ?     How  many- 
have  all?  Ans.    D  has  43590  dollars. 

49.  A  man  bought  three  houses  and  lots  for  15780  dollars, 
and  sold  them  so  as  to  gain  695  dollars  on  each  lot ;  for  how 
much  did  he  sell  them  ?  Ans.   17865  dollars. 

50.  At  the  battle  of  Waterloo,  which  took  place  June  18th, 
1815,  the  estimated  loss  of  the  French  was  40000  men;  of 
the  Prussians,  38000 ;  of  the  Belgians,  8000  ;  of  the  Hano- 
verians, 3500 ;  and  of  the  English,  12000  ;  what  was  the  entire 
loss  of  life  in  this  battle  ? 

51.  The    expenditures    for  educational   purposes   in  New 
England  for  the  year  1850  were  as  follows :  Maine,  380623 
dollars ;    New  Hampshire,  221146  dollars  ;  Vermont,  246604 
dollars ;    Massachusetts,    1424873    dollars  ;    Rhode    Island, 
136729  dollars;  and  Connecticut,  430826  dollars;  what  was 
the  total  expenditure  ?  Ans.    2840801  dollars. 

52.  The    eastern     continent    contains    31000000    square 
miles  ;    the  western   continent,  13750000 ;  Australia,  Green- 
land, and  other  islands,  5250000 ;  what  is  the  entire  area  of 
the  land  surface  of  the  globe  ? 

53.  The  population  of  New  York,  in  1850,  was  515547; 
Boston,    136881;    Philadelphia,   340045;    Chicago,    29963; 
St.  Louis,  77860;  New  Orleans,  116375;  what  was  the  en- 
tire population  of  these  cities  ?  Ans.   1216671. 

54.  The  population  of  the  globe  is  estimated  as  follows : 
North  America,  39257819;  South  America,  18373188;  Eu- 
rope,    265368216  ;    Asia,    630671661  ;    Africa,    61688779  ; 
Oceanica,    23444082 ;   what   is   the  total   population    of  the 
globe  according  to  this  estimate  ?  Ans.   1038803745. 


ADDITION. 


31 


55.  The  railroad  distance  from  New  York  to  Albany  is  144 
miles ;  from  Albany  to  Buffalo,  298 ;  from  Buffalo  to  Cleveland, 
183 ;  from  Cleveland  to  Toledo,  109  ;  from  Toledo  to  Spring- 
field, 365;  and  from  Springfield  to  St.  Louis,  95  miles;  what 
is  the  distance  from  New  York  to  St.  Louis  ? 

56.  A  man  owns  farms  valued  at  56800  dollars ;  city  lots 
valued  at  86760  dollars ;  a  house  worth  12500  dollars,  and 
other  property  to  the  amount  of  6785  dollars ;  what  is  the 
entire  value  of  his  property?  Ans.   162845  dollars. 


(57.) 

15038 

7404 

34971 

30359 

6293 

2875 

16660 

64934 

80901 

7444 

57068 

17255 

32543 

40022 

56063 

33860 

17548 

28944 

16147 

38556 

234882 

3-9058 

152526 

179122 

7626 

1218099 


(58.) 

26881 

12173 

39665 

33249 

6318 

4318 

34705 

80597 

95299 

8624 

53806 

18647 

41609 

35077 

46880 

41842 

26876 

36642 

29997 

44305 

262083 

39744 

169220 

198568 

8735 

1395860 


(59.) 
41919 
19577 
74736 
66768 
12673 
7193 
51365 

155497 

183134 
16845 

111139 
35902 
82182 
75153 

132936 
82939 
44424 
65586 
52839 
83211 

522294 
78861 

353428 

386214 
17005 


(60.) 

93808 

41371 

110525 

102936 

17087 

13251 

112110 

220619 

225255 

68940 

176974 

86590 

149162 

109355 

283910 

112511 

72908 

157672 

86160 

119557 

839398 

117787 

471842 

571778 

41735 


82  SIMPLE   NUMBERS, 


SUBTEACTION. 

MENTAL    EXERCISES. 

43.  1.   A  farmer,  having  14  cows,  sold  6  of  them;  how 
many  had  he  left  ? 

ANALYSIS.    He  had  as  many  left  as  14  cows  less  6  cows,  which 
are  8  cows.    Therefore,  he  had  8  cows  left. 

2.  Stephen,  having  9  marbles,  lost  4  of  them ;  how  many 
had  he  left? 

3.  If  a  man  earn  10  dollars  a  week,  and  spend  6  dollars  for 
provisions,  how  many  dollars  has  he  left  ? 

4.  A  merchant,  having  16  barrels  of  flour,  sells  9  of  them ; 
how  many  has  he  left  ? 

5.  Charles  had  18  cents,  and  gave  10  of  them  for  a  book ; 
how  many  had  he  left  ? 

6.  James  is  17  years  old,  and  his  sister,  Julia,  is  o  years 
younger ;  how  old  is  Julia  ? 

7.  A  grocer,  having  20  boxes  of  lemons,  sold  1 1  boxes ; 
how  many  boxes  had  he  left  ? 

8.  From  a  cistern  containing  25  barrels  of  water,  15  bar- 
rels leaked  out ;  how  many  barrels  remained  ? 

9.  Paid  16  dollars  for  a  coat,  and  7  dollars  for  a  vest; 
how  much  more  did  the  coat  cost  than  the  vest  ? 

10.  How  many  are  18  less  5  ?     17  less  8  ?     12  less  7  ? 

11.  How  many  are  20  less  14  ?     18  less  12  ?     19  less  11  ? 

12.  How  many  are  11  less  3?    16  less  11?    19  less  8? 
20  less  9  ?    22  less  20  ? 

44.  Subtraction  is  the  process  of  taking  one  number  from 
another  equal  to  or  greater  than  itself. 

45.  The  Minuend  is  the  number  to  be  subtracted  from. 

46.  The  Subtrahend  is  the  number  to  be  subtracted. 

Define  subtraction.     Minuend.     Subtrahend. 


SUBTRACTION. 


33 


47.  The  Difference  or  Remainder  is  the  result  obtained 
by  the  process  of  subtraction. 

NOTE.  The  minuend  and  subtrahend  must  be  like  numbers ;  thus, 
5  dollars  from  9  dollars  leave  4  dollars ;  5  apples  from  9  apples  leave  4 
apples ;  but  it  would  be  absurd  to  say  5  apples  from  9  dollars,  or  5 
dollars  from  9  apples. 

48.  The  sign,  — ,  is  called  minus,  which  signifies  less. 
"When  placed  between  two  numbers,  it  denotes  that  the  one 
after  it  is  to  be  taken  from  the  one  before  it.     Thus,  8  —  6  =  2 
is  read  8  minus  6  equals  2,  and  denotes  that  6,  the  subtrahend, 
taken  from  8,  the  minuend,  equals  2,  the  remainder. 

SUBTRACTION    TABLE. 


1  —  I—  0 

2—  2=  0 

3—  3=  0 

4—  4=  0 

2  —  1=  1 

3—  2=  1 

4—  3=  1 

5—  4=  1 

3—  1—  2 

4—  2=  2 

5—  3=  2 

6—  4=  2 

4  —  1—  3 

5—  2=  3 

6—  3=  3 

7—  4=  3 

5—  1  =  4 

6—  2=  4 

7—  3=  4 

8—  4=  4 

6—  1  =  5 

7—  2=  5 

8—  3=  5 

9—  4=  5 

7—  I—  6 

8—  2=  6 

9—  3=  6 

10—  4=  6 

8^7 
HZ   / 

9—  2=  7 

10—  3=  7 

11—  4=  7 

9—  =  8 

10—  2=  8 

11—  3=  8 

12—  4=  8 

10—  =  9 

11—  2=  9 

12—  3=  9 

13—  4=  9 

11—  =10 

12—  2  =  10 

13—  3  =  10 

14—  4  =  10 

12  —  =11 

13—  2=11 

14—  3  =  11 

15—  4  =  11 

13—  =12 

14—  2=12 

15—  3  =  12 

16—  4  =  12 

5—  5=  0 

6—  6=  0 

7—  7=  0 

8—  8=  0 

6—  5=  1 

7—  6=  1 

8—  7=  1 

9—  8=  1 

7  —  5=  2 

8—  6=  2 

9—  7=  2 

10—  8=  2 

8—  5  =  3 

9—  6=  3 

10—  7=  3 

H_  8=  3 

9—  5=  4 

10—  6=  4 

11—  7=  4 

12—  8=  4 

10—5=  5 

11—  6=  5 

12—  7=  5 

13—8=  5 

11—  5  =  6 

12—  6=  6 

13  _  7=  6 

14—8=  6 

12—  5=  7 

13—  6=  7 

14  —  7  =  7 

15—8=  7 

13—5=  8 

14—6=  8 

15—7=  8 

16—8=  8 

14—5=  9 

15—6=  9 

16—  7=  9 

17—  8=  9 

15—  5  =  10 

16—  6  =  10 

17—7  =  10 

18—8  =  10 

16—  5  =  11 

17—  6  =  11 

18—  7  =  11 

19—  8  =  11 

17—5  =  12 

18—  6  =  12 

19—  7  =  12 

20—  8  =  12 

Difference  or  remainder.     What  is  the  sign  of  subtraction  ? 


34 


SIMPLE  NUMBERS. 


9—  9—  0 

10  —  10=  0 

11  —  11=  0 

12  —  12=  0 

10—  9=  1 

11  —  10=  1 

12  —  11=  1 

13  —  12=  1 

11—  9—  2 

12  —  10=  2 

13  —  11=  2 

14  —  12=  2 

12—  9  =  3 

13  —  10=  3 

14  —  11=  3 

15  —  12=  3 

13—  9—  4 

14  —  10=  4 

15  —  11—  4 

16  —  12=  4 

14—  9=  5 

15  —  10=  5 

16  —  11=  5 

17  —  12=  5 

15—  9=  6 

16  —  10=  6 

17  —  11=  6 

18  —  12=  6 

16—  9=  7 

17  —  10=  7 

18  —  11=  7 

19  —  12=  7 

17—  9=  8 

18  —  10=  8 

19  —  11=  8 

20  —  12=  8 

18—  9=  9 

19  —  10=  9 

20—11=  9 

21  —  12=  9 

19—  9=10 

20—10  =  10 

21  —  11  =  10 

22  —  12  =  10 

20—  9  =  11 

21  —  10  =  11 

22  —  11  =  11 

23  —  12  =  11 

21—  9=12 

22—10  =  12 

23  —  11  =  12 

24  —  12  =  12 

CASE   I. 

49.    When  no  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 

1.    From  574  take  323. 


OPERATION. 
Minuend,  574 

Subtrahend,  323 


Remainder, 


251 


ANALYSIS.  We  write  the  less  num- 
ber under  the  greater,  with  units  under 
units,  tens  under  tens,  &c.,  and  draw 
a  line  underneath.  Then,  beginning  at 
the  right  hand,  we  subtract  separately 
each  figure  of  the  subtrahend  from  the 
figure  above  it  in  the  minuend.  Thus,  3  from  4  leaves  1 ,  which  is  the 
difference  of  the  units ;  2  from  7  leaves  5,  the  difference  of  the  tens  ; 
3  from  5  leaves  2,  the  difference  of  the  hundreds.  Hence,  we  have 
for  the  whole  difference,  2  hundreds  5  tens  and  1  unit,  or  251. 


EXAMPLES   FOR    PRACTICE. 


(20 

(3.) 

(4.) 

(5.) 

Minuend, 

876 

676 

367 

925 

Subtrahend, 

334 

415 

152 

213 

Remainder, 

542 

Oaso  1  is  what  ?     Give  explanation. 


SUBTRACTION. 


35 


From 
Take 


(6.) 
876 
523 


(7.) 
732 
522 


(8.) 

987 

782 


(9.) 
498 
178 


Remainders. 

10.  From  3276  take  2143.  1133. 

11.  From  7634  take  3132.  4502. 

12.  From  41763  take  11521.  30242. 

13.  From  18346  take  5215.  13131. 

14.  From  397631  take  175321.  222310. 

15.  Subtract  47321  from  69524.  22203. 

16.  Subtract  16330  from  48673.  32343. 

17.  Subtract  291352  from  895752.  604400. 

18.  Subtract  84321  from  397562.  313241. 

19.  A  farmer  paid  645  dollars  for  a  span  of  horses  and 
a  carriage,  and  sold  them  for  522  dollars ;  how  much  did  he 
lose? 

20.  A  man  bought  a  mill  for  3724  dollars,  and  sold  it  for 
4856  dollars ;  how  much  did  he  gain  ?     Ans.    1132  dollars. 

21.  A  drover  bought  1566  sheep,  and  sold  435  of  them; 
how  many  had  he  left?  Ans.    1131  sheep. 

22.  A  piece  of  land  was  sold  for  2945  dollars,  which  was  832 
dollars  more  than  it  cost ;  what  did  it  cost  ? 

23.  A  gentleman  willed  to  his  son  15768  dollars,  and  to 
his  daughter  4537  dollars ;  how  much  more  did  he  will  to  the 
son  than  to  the  daughter  ?  Ans.    11231  dollars. 

24.  A  merchant  sold  goods  to  the  amount  of  6742  dollars, 
and  by  so  doing  gained  2540  dollars ;  what  did  the  goods  cost 
him  ? 

25.  If  I  borrow  15475  dollars  of  a  person,  and  pay  him 
9050  dollars,  how  much  do  I  still  owe  him  ? 

26.  In  1850  the  white  population  of  the  United  States  was 
19,553,068,  and  the  slave  population  3,204,313 ;   how  much 
was  the  difference? 

27.  The  population  of  Great  Britain  in  1851  was  20,936,468, 
and  of  England  alone,  16,921,888;  what  was  the  difference? 


36  SIMPLE   NUMBERS. 

CASE   II. 

«5O.    When  any  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 
1.   From  846  take  359.  t 

OPERATION.  ANALYSIS.    In  this   example  we 

(7)  (13)  (16)       cannot  take  9  units  from  6  units. 

Minuend,  846         From  the  4  tens  we  take  1  ten,  which 

Subtrahend,         359         equals  10  units,  and  add  to  the  6 

Remainder         ~4     8     7~         un^s>  making  16  units  ;  9  units  from 

16   units  leave  7   units,  which  we 

write  in  the  remainder  in  units'  place.  As  we  have  taken  1  ten 
from  the  4  tens,  3  tens  only  are  left.  We  cannot  take  5  tens  from 
3  tens  ;  so  from  the  8  hundreds  we  take  1  hundred,  which  equals  10 
tens,  and  add  to  the  3  tens,  making  13  tens;  5  tens  from  13 
tens  leave  8  tens,  which  we  write  in  the  remainder  in  tens'  place. 
As  we  have  taken  1  hundred  from  the  8  hundreds,  7  hundreds  only 
are  left ;  3  hundreds  from  7  hundreds  leave  4  hundreds,  which  we 
write  in  the  remainder  in  hundreds'  place,  and  we  have  the  whole 
remainder,  487. 

NOTE.  The  numbers  written  over  the  minuend  are  used  simply  to 
explain  more  clearly  the  method  of  subtracting ;  in  practice  the  pro- 
cess should  be  performed  mentally,  and  these  numbers  omitted. 

The  following  method  is  more  in  accordance  with  prac- 
tice. 

OPERATION.        ANALYSIS.     Since  we  cannot  take  9  units  from  6 
l^s  units,  we  add  10  units  to  6  units,  making  16  units ; 

Jsl  I  9  units  from  16  units  leave  7  units.     But  as  we  have 

added  10  units,  or  1  ten,  to  the  minuend,  we  shall 
359  have  a  remainder  1  ten  too  large,*to  avoid  which,  we 

add  1  ten  to  the  5  tens  in  the  subtrahend,  making  6 
tens.  We  can  not  take  6  tens  from  4  tens ;  so  we  add 
10  tens  to  4,  making  14  tens ;  6  tens  from  14  tens 
leave  8  tens.  Now,  having  added  10  tens,  or  1  hundred,  to  the 
minuend,  we  shall  have  a  remainder  1  hundred  too  large,  unless  we 
add  1  hundred  to  the  3  hundreds  in  the  subtrahend,  making  4  hun- 
dreds ;  4  hundreds  from  8  hundreds  leave  4  hundreds,  and  we  have 
for  the  total  remainder,  487,  the  same  as  before. 

Case  n  is  what  ?    Give  explanation.     Second  explanation, 


SUBTRACTION. 


37 


NOTE.  The  process  of  adding  10  to  the  minuend  is  sometimes  called 
borrowing  10,  and  that  of  adding  1  to  the  next  figure  of  the  subtrahend, 
carrying  one. 

51.  From  the  preceding  examples  and  .illustrations  we 
have  the  following  general 

RULE.  I.  Write  the  less  number  under  the  greater,  placing 
units  of  the  same  order  under  each  other. 

II.  Begin  at  the  right  hand,  and  take  each  figure  of  the  sub- 
trahend from  the  figure  above  it,  and  write  the  result  under- 
neath. 

III.  If  any  figure  in  the  subtrahend  be  greater  than  the  cor- 
responding figure   above  it,  add  10  to  that  upper  figure  be- 
fore subtracting,  and  then  add  1  to  the  next  left  hand  figure  of 
the  subtrahend. 

PROOF.  Add  the  remainder  to  the  subtrahend,  and  if  their 
sum  be  equal  to  the  minuend,  the  work  is  supposed  to  be  right. 


EXAMPLES    FOR   PRACTICE. 


(2.) 

(3.) 

(*•) 

(5.) 

Minuend, 

873 

7432 

1969 

8146 

Subtrahend, 

538 

6711 

1408 

4377 

Remainder, 

335 

(6.) 

(7.) 

(8.) 

(9.) 

gallons. 

bushels. 

miles. 

days. 

From 

3176 

9076 

7320 

5097 

Take 

2907 

4567 

3871 

3809 

(10.) 

(11.) 

(12.) 

(13.) 

dollars. 

rods. 

acres. 

feet. 

From 

76377 

67777 

900076 

767340 

Take 

45761 

46699 

899934 

5039 

What  do  we  mean  hy  borrowing  10  ?    By  carrying  ?    Kule,  first  step  ? 
Second?    Third?    Proof? 


38 


SIMPLE  NUMBERS. 


14.  479  —  382  =  how  many  ?  Ans.   97. 

15.  6593  — 1807  =  how  many?  Ans.   4786. 

16.  17380  —  3417  =  how  many  ?  Ans.    13963. 

17.  80014  — r  43190  =  how  many  ?  Ans.   36824. 

18.  282731  — 90756  — how  many?          Ans.   191975. 

19.  From  234100  take  9970. 

20.  From  345673  take  124799. 

21.  From  4367676  take  256569.  Ans.   4111107. 

22.  From  3467310  take  987631.  Ans.    2479679. 

23.  From  941000  take  5007.  Ans.   935993. 

24.  From  1970000  take  1361111.  Ans.    608889. 

25.  From  290017  take  108045. 

26.  Take  3077097  from  7045 676.  Ans.   3968579. 

27.  Take  9999999  from  60000000.         Ans.   50000001. 

28.  Take  220202  from  4040053.  Ans.   3819851. 

29.  Take  2199077  from  3000001.  Am.   800924. 

30.  Take  377776  from  8000800.  Ans.   7623024. 

31.  Take  501300347  from  1030810040. 

32.  Subtract  nineteen  thousand  nineteen  from  twenty  thou- 
sand ten.  Ans.    991. 

33.  From  one  million  nine  thousand  six  take  twenty  thou- 
sand four  hundred.  Ans.   988606. 

34.  What   is   the   difference   between  two  million   seven 
thousand   eighteen,  and   one   hundred   five   thousand   seven- 
teen? Ans.    1902001. 

35.  What  is  the  difference  between  thirteen  thousand  thir- 
teen, and  eleven  hundred  eleven?  Am.    11902. 

36.  From  fifty  million  take  five  hundred  five. 

37.  A  merchant  bought  5875  bushels  of  wheat,  and  sold 
2976  bushels;    how  many  bushels   remained  in   his  posses- 
sion ?  Am.   2899  bushels. 

38.  The  Declaration  of  Independence  was  published  July 
4,  1776 ;  how  many  years  to  July  4,  1860  ? 

39.  Massachusetts  contains  7800  square  miles,  and  New 
Hampshire  9491  square  miles ;  which  is  the  larger,  and  how 
many  square  miles  ? 


SUBTRACTION.  39 

40.  Sir  Isaac  Newton  was  born  in  the  year  1642,  and  died 
in  1727  ;  to  what  age  did  he  live  ? 

41.  Mont  Blanc,  in  Europe,  is  15,680  feet  high,  and  Chim- 
borazo,  in  America,  is  21,427  feet ;  what  is  the  difference  in 
their  heights  ?  Am.    5747  feet. 

42.  La  Place,  the  celebrated  mathematician  and  philoso- 
pher, was  born  in  1749,  and  died  in  1827 ;  how  long  did  he 
live? 

43.  If  a  company  enter  into   business  with  a  capital  of 
12750  dollars,  and  at  the  end  of  the  year  have  27194  dollars, 
how  much  will  they  have  gained  ?          Ans.    14444  dollars. 

44.  The   population   of  the   United   States   in  1800  was 
5,305,925,  and  in  1850,  23,387,125  ;  what  was  the  increase  in 
50  years?  Ans.    18,081,200. 

45.  The  population   of  the  United  States  by  the   census 
of  1850  was  23,387,125;  it  is  estimated  that  in  1860  it  wjll 
be  31,095,535 ;  what  is   the   estimated  increase   for  the   10 
years?  Ans.  7,708,410. 

46.  According  to  the  same  rate  of  increase,  the  population 
of  the  United  States  in  1870  will  be  40,617,708  ;  what  will 
be  the  increase  of  population  from  1850  ?      Ans.   17,230,583. 

47.  According  to  the  same  rate  of  increase,  in  1900  the 
population  will  be  101553377 ;  what  will  be  the  increase  from 
1850?  Ans.   78166252. 


EXAMPLES    COMBINING   ADDITION    AND    SUBTRACTION. 


1.  A  merchant  gave  his  note  for  5200  dollars.  He 
paid  at  one  time  2500  dollars,  and  at  another  175  dollars; 
what  remained  due  ?  Ans.  2525  dollars. 

2.  A  traveler  who  was  1300  miles  from  home,  traveled 
homeward  235  miles  in  one  week,  in  the  next  275  miles,  in  the 
next  325  miles,  and  in  the  next  280  miles  ;  how  far  had  he 
still  to  go  before  he  would  reach  home?       Ans.    185  miles. 

3.  A  man  deposited  in  bank  8752  dollars  ;  he  drew  out  at 
one  time  4234  dollars,  at  another  1700  dollars,  at  another  962 


40  SIMPLE  NUMBERS. 

dollars,  and  at  another  49  dollars  ;  how  much  had  he  remain- 
ing in  bank  ?  Ans.    1807  dollars. 

4.  A  man  bought  a  farm  for  4765  dollars,  and  paid  750 
dollars  for  fencing  and  other  improvements  ;  he  then  sold  it  for 
384  dollars  less  than  it  cost  him ;  how  much  did  he  receive 
for  it?  Ans.   5131  dollars. 

5.  A  forwarding  merchant  had  in  his  warehouse  7520  bar- 
rels of  flour;  he  shipped  at  one  time  1224  barrels,  at  another 
time   1500  barrels,  and  at  another  time    1805  barrels;  how 
many  barrels  remained  ? 

6.  A   had  450  sheep,  B  had   175  more  than  A,  and  C 
had  as  many  as  A  and  B  together  minus  114;  how  many 
sheep  had  C  ?  Ans.    961  sheep. 

7.  A  farmer  raised  1575  bushels  of  wheat,  and  900  bushels 
of  corn.     He  sold  807  bushels  of  wheat,  and  391  bushels  of 
corn  to  A,  and  the  remainder  to  B ;  how  much  of  each  did 
he  sell  to  B  ?    Ans.   768  bushels  of  wheat,  and  509  of  corn. 

8.  A  man  traveled  6784  miles ;   2324  miles  by  railroad, 
1570  miles  in  a  stage  coach,  450  miles  on  horseback,  175 
miles  on  foot,  and  the  remainder  by  steamboat ;  how  many 
miles  did  he  travel  by  steamboat  ?  Ans.    2265  miles. 

9.  Three  persons  bought  a  hotel  valued  at  35680  dollars. 
The  first  agreed  to  pay  7375  dollars,  the  second  agreed  to 
pay  twice  as  much,  and  the  third  the  remainder ;  how  much 
was  the  third  to  pay  ?  Ans.    13555  dollars. 

10.  Borrowed  of  my  neighbor  at  one  time  750  dollars,  at 
another  time  379  dollars,  and  at  another  450  dollars.     Having 
paid  him  1000  dollars,  how  much  do  I  still  owe  him? 

Ans.    579  dollars. 

11.  A  man  worth  6709  dollars,  received  a  legacy  of  3000 
dollars.     He  spent  4379  dollars  in  traveling;  how  much  had 
he  left  ? 

12.  In   1850  the  number  of  white  males  in  the    United 
States  was  10026402,  and    of   white   females    9526666;   of 
these,  8786968    males,    and    8525565    females    were    native 
born;  how  many  of  both  were  foreign  born?    Ans.    2240535. 


MULTIPLICATION.  41 


MULTIPLICATION. 

MENTAL    EXERCISES. 

1.   What  will  4  pounds  of  sugar  cost,  at  8  cents  a 
pound  ? 

ANALYSIS.  Four  pounds  will  cost  as  much  as  the  price,  8  cents 
taken  4  times  ;  thus,  8  -j-  8  +  8  -\-  8  —  32.  But  instead  of  adding, 
we  may  say,  —  since  one  pound  costs  8  cents,  4  pounds  will  cost  4 
times  8  cents,  or  32  cents. 

2.  If  a  ream  of  paper  cost  3  dollars,  what  will  2  reams 
cost? 

3.  At   7    cents  a   quart,  what  will   4  quarts  of  cherries 
cost? 

4.  At  12  dollars  a  ton,  what  will  3  tons  of  hay  cost?  4 
tons?   5  tons? 

5.  There  are  7  days  in  1  week ;  how  many  days  in  6  weeks  ? 
in  8  weeks  ? 

6.  What  will  9  chairs  cost,  at  10  shillings  apiece  ? 

7.  If  Henry  earn  12  dollars  in  1  month,  how  much  can  he 
earn  in  5  months  ?  in  7  months  ?  in  9  months  ? 

8.  What  will  11  dozen  of  eggs  cost,  at  9  cents  a  dozen?  at 
10  cents?  at  12  cents? 

9.  When  flour  is  7  dollars  a  barrel,  how  much  must  be 
paid  for  7  barrels  ?  for  9  barrels  ?  for  12  barrels  ? 

10.  At  9  dollars  a  week,  what  will  4  weeks'  board  cost  ? 
7  weeks'  ?  9  weeks'  ? 

11.  If  I  deposit  12  dollars  in  a  savings  bank  every  month, 
how  many  dollars  will  I  deposit  in  6  months  ?  in  8  months  ? 
in  9  months  ? 

12.  At  9  cents  a  foot,  what  will  4  feet  of  lead  pipe  cost  ? 
7  feet?   10  feet? 

13.  When  hay  is  8  dollars  a  ton,  how  much  will  3  tons 
cost  ?  4  tons  ?  7  tons  ?  9  tons  ?   11  tons  ? 

D* 


42 


SIMPLE  NUMBERS. 


14.  What  will  be  the  cost  of  11  barrels  of  apples,  at  2  dol- 
lars a  barrel  ?  at  3  dollars  ? 

15.  At  1 0  cents  a  pound,  what  will  9  pounds  of  sugar  cost  ? 
11  pounds  ?   12  pounds  ? 

54.  Multiplication  is  the  process  of  taking  one  of  two 
given  numbers  as  many  times  as  there  are  units  in  the  other. 

55.  The  Multiplicand  is  the  number  to  be  taken. 

56.  The   Multiplier  is   the   number  which  shows   how 
many  times  the  multiplicand  is  to  be  taken. 

57.  The  Product  is  the  result  obtained  by  the  process  of 
multiplication. 

58.  The  Factors  are  the  multiplicand  and  multiplier. 

NOTES.  1.  Factors  are  producers,  and  the  multiplicand  and  mul- 
tiplier are  called  factors  because  they  produce  the  product. 

2.  Multiplication  is  a  short  method  of  performing  addition  when 
the  numbers  to  be  added  are  equal. 

59.  The  sign,  X>  placed  between  two  numbers,  denotes 
that  they  are  to  be  multiplied  together ;  thus  9  X  6  =  54,  is 
read  9  times  6  equals  54. 


MULTIPLICATION   TABLE. 


IX    1  =   1 

2X    1=   2 

3X    1—   3 

4X    1=   4 

IX    2  =   2 

2X    2  =   4 

3X    2=   6 

4X    2=   8 

X    3=   3 

2X    3  =   6 

3X    3=   9 

4x    3  =  12 

X    4=   4 

2X    4=   8 

3X    4  =  12 

4X    4  =  16 

X    5=   5 

2X    5  =  10 

3X    5  =  15 

4x    5  =  20 

X    6=   6 

2X    6  =  12 

3X    6=18 

4x    6  =  24 

X    7  =   7 

2X    7  =  14 

3X    7  =  21 

4x    7  =  28 

X    8=   8 

2X    8  =  16 

3X    8  =  24 

4X    8  =  32 

IX    9=   9 

2X    9  =  18 

3X    9  =  27 

4  X    9  =  36 

1  X  10  —  10 

2  X  10  =  20 

3X  10  =  30 

4X  10  =  40 

1  X  11  =  11 

2  X  11  —  22 

3  X  11  =  33 

4  X  11=44 

IX  12=12 

2X  12  =  24 

3  X  12  =  36 

4  X  12  =  48 

Define  multiplication.  Multiplicand.  Multiplier.  Product.  Fac- 
tors. Multiplication  is  a  short  method  of  what  ?  What  is  the  sign  of 
multiplication  ? 


MULTIPLICATION. 


43 


5X    1=    5 

6X    1=   6 

7X    1=   7 

8X    1=   8 

5X    2  =  10 

6X    2  =  12 

7X    2  =  14 

8X    2  =  16 

5X    3  =  15 

6X    3  =  18 

7X    3  =  21 

8  X    3  =  24 

5  X    4  =  20 

6X    4  =  24 

7  X    4  =  28 

8X    4  =  32 

5X    5  =  25 

6X    5  =  30 

7X    5  =  35 

8X    5  =  40 

5  X    6  =  30 

6  X    6  =  36 

7X    6  =  42 

8X    6  =  48 

5X    7  =  35 

6  X    7  =  42 

7X    7  =  49 

8X   7  =  56 

5X    8  =  40 

6  X    8  =  48 

7X    8  =  56 

8  X    8  =  64 

5X    9  =  45 

6X    9  =  54 

7X    9  =  63 

8X    9  =  72 

5  X  10  =  50 

6X  10  =  60 

7X  10  =  70 

8  X  10  =  80 

5X  11  =  55 

6  X  11  =  66 

7X  11  =  77 

8  X  11  =  88 

5  X  12  =  60 

6X  12  =  72 

7X  12  =  84 

8X  12  =  96 

9X    1=     9 

10  X   1=    10 

11  X   1=    11 

12  X  1=    12 

9X    2=    18 

10  X  2  =   20 

11  X  2=   22 

12  X  2=   24 

9X    3=   27 

10  X  3=   30 

11  X  3=   33 

12  X  3=   36 

9  X    4  =   36 

10  X  4  =   40 

11  X  4=   44 

12  X  4=   48 

9  X    5  =   45 

10  X  5  =   50 

11  X  5  —   55 

12  X  5=    60 

9  X    6  =   54 

10  X  6=   60 

11  X  6=   66 

12  X  6=    72 

9X    7=   63 

10  X  7  =   70 

11  X  7=   77 

12  X   7=   84 

9X    8=    72 

10  X  8  =   80 

11  X  8=   88 

12  X  8=   96 

9  X    9=   81 

10  X  9=   90 

11  X  9=   99 

12  X  9  =  108 

9  X  10  =   90 

10X10  =  100 

11  X10  =  110 

12X10  =  120 

9X  11=   99 

10x11  =  110 

11  XH  =  121 

12X11  =  132 

9X  12  =  108 

10X12  =  120 

11  X12  =  132 

12X12  =  144 

OPERATION. 


CASE    I. 

6O.   When  the  multiplier  consists  of  one  figure. 
1.   Multiply  374  by  6. 

ANALYSIS.  In  this  example  it  is 
Lri  required  to  take  374  six  times.  If  we 
l||  take  the  units  of  each  order  6  times, 
we  shall  take  the  entire  number  6 
times.  Therefore,  writing  the  multi- 
plier under  the  unit  figure  of  the  mul- 
tiplicand, we  proceed  as  follows:  6 
times  4  units  are  24  units  ;  6  times  7 
tens  are  42  tens  ;  6  times  3  hundreds 
are  18  hundreds;  and  adding  these 
partial  products,  we  obtain  the  entire 
product,  2244. 


Multiplicand, 
Multiplier, 

units, 
tens, 
hundreds, 

Product, 

374 
6 

24 
42 
18 

2244 

Case  I  is  what  ?     Give  explanation. 


44  SIMPLE  NUMBERS. 

The  operation  in  this  example  may  be  performed  in  another 
way,  which  is  the  one  in  common  use. 

OPERATION.  ANALYSIS.  Writing  the  numbers  as  before,  we 
374  begin  at  the  right  hand  or  unit  figure,  and  say:  6 
times  4  units  are  24  units,  which  is  2  tens  and  4 
units;  write  the  4  units  in  the  product  in  units' 
place,  and  reserve  the  2  tens  to  add  to  the  next  prod- 
uct ;  6  times  7  tens  are  42  tens,  and  the  two  tens  re- 
served in  the  last  product  added,  are  44  tens,  which  is  4  hundreds 
and  4  tens ;  write  the  4  tens  in  the  product  in  tens'  place,  and  reserve 
the  4  hundreds  to  add  to  the  next  product ;  6  times  3  hundreds  are 
18  hundreds,  and  4  hundreds  added  are  22  hundreds,  which,  being 
written  in  the  product  in  the  places  of  hundreds  and  thousands, 
gives,  for  the  entire  product,  2244. 

Gl.  The  unit  value  of  a  number  is  not  changed  by  re- 
peating the  number.  As  the  multiplier  always  expresses 
times,  the  product  must  have  the  same  unit  value  as  the  mul- 
tiplicand. But,  since  the  product  of  any  two  numbers  will  be 
the  same,  whichever  factor  is  taken  as  a  multiplier,  either 
factor  may  be  taken  for  the  multiplier  or  multiplicand. 

NOTE.  In  multiplying,  learn  to  pronounce  the  partial  results,  as  in 
addition,  without  naming  the  numbers  separately;  thus,  in  the  last 
example,  instead  of  saying  6  times  4  are  24,  6  times  7  are  42  and  2  to 
carry  are  44,  6  times  3  are  18  and  4  to  carry  are  22,  pronounce  only 
the  results,  24,  44,  22,  performing  the  operations  mentally.  This  will 
greatly  facilitate  the  process  of  multiplying. 


EXAMPLES 

FOB  PRACTICE. 

Multiplicand, 

(2.) 
7324 

(3.) 

6812 

(4.) 
34651 

Multiplier, 
Product, 

4 

6 

5 

29296 

40872 

173255 

(5.) 

82456 

(6.) 
92714 

(?•) 
28093 

(8.) 
46247 

3 

7 

8 

9 

Second  explanation.     Repeating  a  number  has  what  effect  on  the 
unit  value  ?    The  product  must  be  of  the  same  kind  as  what  ? 


MULTIPLICATION.  45 

9.   Multiply    32746  by  5.  Ans.      163730. 

10.  Multiply  840371  by  7.  Ans.   5882597. 

11.  Multiply  137629  by  8.  Ans.    1101032. 

12.  Multiply    93762  by  3.  Ans.     281286. 

13.  Multiply  543272  by  4.  Ans.   2173088. 

14.  Multiply  703164  by  9.  Ans.   6328476. 

15.  What  will  be  the  cost  of  344  cords  of  wood  at  4  dol- 
lars a  cord?  Ans.   1376. 

16.  How  much  will  an  army  of  7856  men  receive  in  one 
week,  if  each  man  receive  6  dollars  ?      Ans.   47136  dollars. 

17.  In  one  day  are  86400  seconds ;  how  many  seconds  in 
7  days  ?  Ans.  604800  seconds. 

18.  What  will  7640  bushels  of  wheat  cost,  at  9  shillings  a 
bushel  ?  Ans.    68760  shillings. 

19.  At  5  dollars  an  acre,  what  will  2487  acres  of  land 
cost?  Ans.    12435  dollars. 

20.  In  one  mile  are  5280  feet ;  how  many  feet  in  8  miles  ? 

Ans.   42240  feet. 

CASE   II. 

62.    When  the  multiplier  consists  of  two  or  more 
figures. 

1.  Multiply  746  by  23. 

OPERATION.  ANALYSIS.      Writ- 

Multiplicand,  746  ing  tne    multiplicand 

Multiplier,  23  and  multiplier  as  in 

Case  I,  we  first  mul- 

2238      »$g£Sr*         tiply  each  figure  in  the 
1492       2o|{gf«atnh|.mul-         multiplicand    by   the 

Product,  17158     »{g3SEf™'-         ™t  figure  of  the  mul- 

tiplier,  precisely  as  m 

Case  I.  We  then  multiply  by  the  2  tens.  2  tens  times  6  units,  or  6 
times  2  tens,  are  12  tens,  equal  to  1  hundred,  and  2  tens ;  we  place  the 
2  tens  under  the  tens  figure  in  the  product  already  obtained,  and  add 
the  1  hundred  to  the  next  hundreds  produced.  2  tens  times  4  tens 
are  8  hundreds,  and  the  1  hundred  of  the  last  product  added  are  9 
hundreds ;  we  write  the  9  in  hundreds'  place  in  the  product.  2  tens 

Case  II  is  what  ?     Give  explanation. 


46  SIMPLE    NUMBERS. 

times  7  hundreds  are  14  thousands,  equal  to  1  ten  thousand  and  4 
thousands,  which  we  write  in  their  appropriate  places  in  the  product. 
Then  adding  the  two  products,,  we  have  the  entire  product,  17158. 

NOTES.  1.  When  the  multiplier  contains  two  or  more  figures,  the 
several  results  obtained  by  multiplying  by  each  figure  are  called  partial 
products. 

2.  When  there  are  ciphers  between  the  significant  figures  of  the 
multiplier,  pass  over  them,  and  multiply  by  the  significant  figures  only. 

B3.  From  the  preceding  examples  and  illustrations  we 
deduce  the  following  general 

RULE.  I.  Write  the  multiplier  under  the  multiplicand,  placing 
units  of  the  same  order  under  each  other. 

II.  Multiply  the  multiplicand  by  each  figure  of  the  multi- 
plier successively,  beginning  with  the  unit  figure,  and  write  the 
first  figure  of  each  partial  product  under  the  figure  of  the  mul- 
tiplier used,  writing  down  and  carrying  as  in  addition. 

III.  If  there  are  partial  products,  add  them,  and  their  sum 
will  be  the  product  required. 

G4:.  PROOF.  1.  Multiply  the  multiplier  by  the  multipli- 
cand, and  if  the  product  is  the  same  as  the  first  result,  the 
work  is  correct.  Or, 

2.  Multiply  the  multiplicand  by  the  multiplier  diminished 
by  1,  and  to  the  product  add  the  multiplicand ;  if  the  sum  be 
the  same  as  the  product  by  the  whole  of  the  multiplier,  the 
work  is  correct. 


Multiply 
By 

Ans. 

EXAMPLES 

(2.) 
4732 
36 

FOK  PRACTICE. 

(3.) 

8721 
47 

(4.) 
17605 
204 

28392 
14196 

61047 
34884 

70420 
35210 

170352 

409887 

3591420 

What  are  partial  products  ?  When  there  are  ciphers  in  the  multi- 
plier, how  proceed  ?  llule,  first  step  ?  Second  ?  Third  ?  Proof, 
first  method  ?  Second  ? 


MULTIPLICATION.  47 


(5.)  (6.) 

7648  81092 

328  194 


8.  How  many  yards  of  linen  in  759  pieces,  each  piece  con- 
taining 25  yards  ?  Am.    18975  yards. 

9.  Sound  is  known  to  travel  about  1142  feet  in  a  second  of 
time  ;  how  far  will  it  travel  in  69  seconds  ? 

10.  A  man  bought  36  city  lots,  at  475  dollars  each ;  how 
mucft  did  they  all  cost  him  ?  Ans.    17100  dollars. 

11.  What  would  be  the  value  of  867  shares  of  railroad 
stock,  at  97  dollars  a  share  ?  Ans.   84099  dollars. 

12.  How  many  pages   in    3475    books,  if  there  be   362 
pages  in  each  book  ?  Ans.    1257950  pages. 

13.  In  a  garrison  of  4507  men,  each  man  receives  annually 
208  dollars ;  how  much  do  they  all  receive  ? 

14.  Multiply  7198  by  216.  Ans.    1554768. 

15.  Multiply  31416  by  175.  Ans.    5497800. 

16.  Multiply  7071  by  556.  Ans.    3931476. 

17.  Multiply  75649  by  579.  Ans.  43800771. 

18.  Multiply  15607  by  3094.  Ans.  48288058. 

19.  Multiply  79094451  by  76095.    Ans.    6018692248845. 

20.  Multiply  live  hundred  forty  thousand  six  hundred  nine, 
by  seventeen  hundred  fifty.  Ans.    946065750. 

21.  Multiply  four  million  twenty-five  thousand  three  hun- 
dred ten,  by  seventy-five  thousand  forty-six. 

Ans.  302083414260. 

22.  Multiply  eight  hundred  seventy-seven  million  five  hun- 
dred ten  thousand  eight  hundred  sixty-four,  by  five  hundred 
forty-five  thousand  three  hundred  fifty-seven. 

Ans.  478556692258448. 

23.  If  one  mile  of  railroad  require  116  tons  of  iron,  worth 
65  dollars  a  ton,  what  will  be  the  cost  of  sufficient  iron  to 
construct  a  road  128  miles  in  length?       Ans.  965120  dollars. 


48  SIMPLE  NUMBERS. 


CONTRACTIONS. 
CASE   I. 

When  the  multiplier  is  a  composite  number. 

A  Composite  Number  is  one  that  may  be  produced  by 
multiplying  together  two  or  more  numbers  ;  thus,  18  is  a  com- 
posite number,  since  6X3  =  18;  or,  9  X  2  =  18  ;  or,  3  X 
3X  2  =  18. 

66.  The  Component  Factors  of  a  number  are  the  sev- 
eral numbers  which,  multiplied  together,  produce  the  given 
number  ;  thus,  the  component  factors  of  20  are  10  and  2, 
(10  X  2  =  20  ;)  or,  4  and  5,  (4  X  5  =  20  ;)  or,  2  and  2  and 
5,  (2  X  2  X  5  =  20.) 

NOTE.  The  pupil  must  not  confound  the  factors  with  the  parts  of  a 
number.  Thus,  the  factors  of  which  12  is  composed,  are  4  and  3, 
(4X3=12  ;)  while  the  parts  of  which  12  is  composed  are  8  and  4, 
(8  -f-  4  =  12,)  or  10  and  2,  (10  +  2  =  12.)  The  factors  are  multiplied, 
while  the  parts  are  added,  to  produce  the  number. 

1.    What  will  32  horses  cost,  at  174  dollars  apiece  ? 

OPERATION.  ANALYSIS.    The  fac- 

Multiplicand,  174    COSt  of  1  llOFSC.          tors    of    32    are    4    and 

1st  factor,  4  8-     If  we  multiply  the 

cost  of  1  horse  by  4, 

696  cost  of  4  horses.      we  obtain  the  cost  of  4 
*2d  factor,  8  horses  ;  and  by  multi- 


Product,  5568  cost  of  32  horses.  ft  cost  _°f  .4 

horses  by  8,  we  obtain 

the  cost  of  8  times  4  horses,  or  32  horses,  the  number  bought. 
67'.    Hence  we  have  the  following 

RULE.  I.  Separate  the  composite  number  into  two  or  more 
factors. 

II.   Multiply  the  multiplicand  by  one  of  these  factors,  and 

What  are  contractions?  Case  I  is  what?  Define  a  composite 
number.  Component  factors.  What  caution  is  given  ?  Give  ex- 
planation. Rule,  first  step  ?  Second  ? 


MULTIPLICATION.  49 

that  product  "by  another,  and  so  on  until  all  the  factors  have 
been  used  successively ;  the  last  product  will  be  the  product  re- 
quired, 

NOTE.  The  product  of  any  number  of  factors  will  be  the  same  in 
whatever  order  they  are  multiplied.  Thus,  4  X  3  X  5  =  60,  and 
5X  4X  3  =  60. 

EXAMPLES    FOR   PRACTICE. 

2.  Multiply  3472  by  48  =  6  X  8.  Ans.    166656. 

3.  Multiply  14761  by  64  =  8  X  8. 

4.  Multiply  87034  by  81  =  3  X  3  X  9.      Ans.    7049754. 

5.  Multiply  47326  by  120  =  6  X  5  X  4. 

6.  Multiply  60315  by  96.  Ans.    5790240. 

7.  Multiply  291042  by  125.  Ans.   36380250. 

8.  If  a  vessel  sail  436  miles  in  1  day,  how  far  will  she  sail 
in  56  days  ?  Ans.    24416  miles. 

9.  How  much  will  72  acres  of  land  cost,  at  124  dollars  an 
acre  ?  Ans.   8928  dollars. 

10.  There  are  5280  feet  in  a  mile;  how  many  feet  in  84 
miles  ?  Ans.   443520  feet. 

11.  What  will  120  yoke  of  cattle  cost,  at  125  dollars  a 
yoke  ? 

CASE    II. 

68.  When  the  multiplier  is  10,  100,  1000,  &c. 

If  we  annex  a  cipher  to  the  multiplicand,  each  figure  is  re- 
moved one  place  toward  the  left,  and  consequently  the  value  of 
the  whole  number  is  increased  tenfold,  (33.)  If  two  ciphers 
are  annexed,  each  figure  is  removed  two  places  toward  the 
left,  and  the  value  of  the  number  is  increased  one  hundred 
fold ;  and  every  additional  cipher  increases  the  value  tenfold. 

69.  Hence  the  following 

RULE.  Annex  as  many  ciphers  to  the  multiplicand  as  there 
are  ciphers  in  the  multiplier ;  the  number  so  formed  will  he 
the  product  required. 

Case  II  is  what  ?     Give  explanation.     Rule  ? 
E    . 


50  SIMPLE   NUMBERS. 

EXAMPLES    HOR    PRACTICE. 

1.  Multiply  347  by  10.  Ans.    3470. 

2.  Multiply  4731  by  100.  Ans.   473100. 

3.  Multiply  13071  by  1000. 

4.  Multiply  89017  by  10000. 

5.  If  1  acre  of  land  cost  36  dollars,   what   will  10  acres 
cost?  Ans.   3 60  dollars. 

6.  If  1  bushel  of  corn  cost  65  cents,  what  will  1000  bushels 
cost  ?  Ans.    65000  cents. 

CASE    III. 

7O.  When  there  are  ciphers  at  the  right  hand  of 
one  or  both  of  the  factors. 

1.   Multiply  1200  by  60. 

OPERATION.  ANALYSIS.    Both  multiplicand  and 

Multiplicand,         1200  multiplier  may  be  resolved  into  their 

Multi  lier  60  component  factors  ;  1200  into  12  and 

100,  and  60  into  6  and  10.     If  these 

Product,  72000          several  factors  be  multiplied  together 

they  will  produce  the  same  product  as 

the  given  numbers,  (67.)  Thus,  12  X  6  =  72,  and  72  X  100  = 
7200,  and  7200  X  10  =  72000,  which  is  the  same  result  as  in  the 
operation.  Hence  the  following 

RULE.  Multiply  the  significant  figures  of  the  multiplicand 
by  those  of  the  multiplier,  and  to  the  product  annex  as  many 
ciphers  as  there  are  ciphers  on  the  right  of  both  factors. 

EXAMPLES    FOR    PRACTICE. 

(2.)  (3.) 

Multiply     4720  10340000 

By         340000  105000 


1888  5170 

1416  1034 


1604800000  1085700000000 


Case  III  is  what  ?     Give  explanation.    Rule. 


MULTIPLICATION.  51 

4.  Multiply  70340  by  800400.        Am.  56300136000. 

5.  Multiply  3400900  by  207000.    Ans.  703986300000. 

6.  Multiply  634003000  by  40020.   Ans.  25372800060000. 

7.  Multiply  10203070  by  50302000. 

Ans.  513234827140000. 

8.  Multiply  30090800  by  600080.  Ans.   18056887264000. 

9.  Multiply  eighty  million  seven  thousand  six  hundred,  by 
eight  million  seven  hundred  sixty.      Ans.   640121605776000. 

10.  Multiply  fifty  million  ten  thousand  seventy,  by  sixty- 
four  thousand.  Ans.   3200644480000. 

11.  Multiply  ten  million  three  hundred  fifty  thousand  one 
hundred,  by  eighty  thousand  nine  hundred. 

Ans.  837323090000. 

12.  There  are  296  members  of  Congress,  and  each  one  re- 
ceives a  salary  of  3000  dollars  a  year ;  how  much  do  they  all 
receive  ? 

EXAMPLES    COMBINING    ADDITION,    SUBTRACTION,   AND 
MULTIPLICATION. 

1.  Bought  45  cords  of  wood  at  4  dollars  a  cord,  and  9  loads 
of  hay  at  13  dollars  a  load;  what  was  the  cost  of  the  wood 
and  hay  ?  Ans.   297  dollars. 

2.  A  merchant  bought  6  hogsheads  of  sugar  at  31  dollars 
a  hogshead,  and  sold  it  for  39  dollars  a  hogshead ;  how  much 
did  he  gain? 

3.  Bought  288  barrels  of  flour  for  1875  dollars,  and  sold 
the  same  for  9  dollars  a  barrel ;  how  much  was  the  gain  ? 

Ans.   Ill  dollars. 

4.  If  a  young  man  receive  500  dollars  a  year  salary  and 
pay  240  dollars  for  board,  125  dollars  for  clothing,  75  dollars 
for  books,  and  50  dollars  for  other  expenses,  how  much  will 
he  have  left  at  the  end  of  the  year?  Ans.    10  dollars. 

5.  A  farmer  sold   184  bushels  of  wheat  at  2  dollars   a 
bushel,  for  which  he  received  67  yards  of  cloth  at  4  dollars  a 
yard,  and  the  balance  in  groceries ;  how  much  did  his  gro- 
ceries cost  him  ? 


52  SIMPLE   NUMBERS. 

6.  A  sold  a  farm  of  320  acres  at  36  dollars  an  acre ;  B 
sold  one  of  244  acres  at  48  dollars  an  acre ;  \vhich  received 
the  greater  sum,  and  how  much?          Ans.    B,  192  dollars. 

7.  Two  persons  start  from  the  same  point  and  travel  in 
opposite  directions,  one  at  the  rate  of  35  miles  a  day,  and  the 
other  29  miles  a  day ;  how  far  apart  will  they  be  in  16  days? 

Ans.    1024  miles. 

8.  A  merchant  tailor  bought  14  bales  of  cloth,  each  bale 
containing  26  pieces,  and  each  piece  43  yards;  how  many 
yards  of  cloth  did  he  buy?  Ans.    15652  yards. 

9.  If  a  man  have  an  income  of  3700  dollars  a  year,  and  his 
daily  expenses  be  4  dollars ;  what  will  he  save  in  a  year,  or 
365  days  ?  Ans.   2240  dollars. 

10.  A  man  sold  three  houses  ;  for  the  first  he  received 
2475  dollars,  for  the  second  840  dollars  less  than  he  received 
for  the  first,  and  for  the  third  as  much  as  for  the  other  two ; 
how  much  did  he  receive  for  the  three  ?     Ans.   8220  dollars. 

11.  A  man  sets  out  to  travel  from  Albany  to  Buffalo,  a 
distance  of  336  miles,  and  walks  28  miles  a  day  for  10  days ; 
how  far  is  he  from  Buffalo  ? 

12.  Mr.  C  bought  14  cows  at  23  dollars  each,  7  horses  at 
96  dollars  each,  34  oxen  at  57  dollars  each,  and  300  sheep  at 
2  dollars  each ;   he   sold   the  whole  for  3842  dollars  ;    how 
much  did  he  gain  ?  Ans.   310  dollars. 

13.  A  drover  bought  164  head  of  cattle  at  36  dollars  a 
head,  and  850  sheep  at  3  dollars  a  head ;  how  much  did  he 
pay  for  all  ? 

14.  A  banker  has  an  income  of  14760  dollars  a  year;  he 
pays  1575  dollars  for  house  rent,  and  four  times  as  much  for 
family  expenses  ;  how  much  does  he  save  annually  ? 

Ans.    6885  dollars. 

15.  A  flour  merchant  bought  936  barrels  of  flour  at  9  dol- 
lars a  barrel;  he  sold  480  barrels  at  10  dollars  a  barrel,  and 
the  remainder  at  8  dollars  a  barrel ;  how  much  did  he  gain  or 
lose  ?  Ans.    Gained  24  dollars. 


DIVISION.  53 


DIVISION. 

MENTAL    EXERCISES. 

71 .    1.   How  many  hats,  at  4  dollars  apiece,  can  be  bought 

for  20  dollars  ? 

« 

ANALYSIS.  Since  4  dollars  will  buy  one  hat,  20  dollars  will  buy 
as  many  hats  as  4  is  contained  times  in  20,  which  is  5  times.  There- 
fore, 5  hats,  at  4  dollars  apiece,  can  be  bought  for  20  dollars. 

2.  A  man  gave  1 6  dollars  for  8  barrels  of  apples ;  what 
was  the  cost  of  each  barrel  ? 

3.  If  1  cord  of  wood  cost  3  dollars,  how  many  cords  can 
be  bought  for  15  dollars  ? 

4.  At  6  shillings  a  bushel,  how  many  bushels  of  corn  can 
be  bought  for  24  shillings  ? 

5.  When  flour  is  6  dollars  a  barrel,  how  many  barrels  can 
be  bought  for  30  dollars  ? 

6.  If  a  man  can  dig  7  rods  of  ditch  in  a  day,  how  many 
days  will  it  take  him  to  dig  28  rods  ? 

7.  If  an  orchard  contain  56  trees,  and  7  trees  in  a  row, 
how  many  rows  are  there  ? 

8.  Bought  6  barrels  of  flour  for  42  dollars  ;  what  was  the 
cost  of  1  barrel  ? 

9.  If   a   farmer   divide    21    bushels   of   potatoes   equally 
among  7  laborers,  how  many  bushels  will  each  receive  ? 

10.  How  many  oranges  can  be  bought  for  27  cents,  at  3 
cents  each? 

11.  A  farmer  paid  35  dollars  for  sheep,  at  5  dollars  apiece ; 
how  many  did  he  buy  ? 

12.  How  many  times  4  in  28  ?  in  16  ?  in  36  ? 

13.  How  many  times  8  in  40  ?  in  56  ?  in  64  ? 

14.  How  many  times  9  in  36  ?  in  63  ?  in  81  ? 

15.  How  many  times  7  in  49  ?  in  70  ?  in  84  ? 

E* 


54 


SIMPLE  NUMBERS. 


7!3.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another. 

73.  The  Dividend  is  the  number  to  be  divided. 

74.  The  Divisor  is  the  number  to  divide  by. 

75.  The  Quotient  is  the  result  obtained  by  the  process  of 
division,  and  shows  how  many  times  the  divisor  is  contained 
in  the  dividend. 

NOTES.  1.  When  the  dividend  does  not  contain  the  divisor  an  exact 
number  of  times,  the  part  of  the  dividend  left  is  called  the  remainder, 
and  it  must  be  less  than  the  divisor. 

2.  As  the  remainder  is  always  a  part  of  the  dividend,  it  is  always 
of  the  same  name  or  kind. 

3.  When  there  is  no  remainder  the  division  is  said  to  be  complete. 

7O.  The  sign,  -j-,  placed  between  two  numbers,  denotes 
division,  and  shows  that  the  number  on  the  left  is  to  be  divided 
by  the  number  on  the  right.  Thus,  20  -f-  4  =  5,  is  read,  20 
divided  by  4  is  equal  to  5. 

Division  is  also  indicated  by  writing  the  dividend  above,  and 

12 
the  divisor  below  a  short  horizontal  line ;  thus,  —  =  4,  shows 

that  12  divided  by  3  equals  4. 


DIVISION    TABLE. 


1- 

-| 

2- 

-  2=  1 

3- 

-  3=  1 

4- 

-  4=  1 

2- 

=  2 

4- 

-  2=  2 

6- 

-3=2 

8- 

-  4=  2 

3- 

—  3 

6- 

-  2=  3 

9- 

-  3=  3 

12- 

-  4=  3 

4- 

=  4 

8- 

-  2=  4 

12- 

-  3=  4 

16- 

-  4—  4 

5- 

—  5 

10- 

-  2=  5 

15- 

-  3=  5 

20- 

-4—5 

6- 

=  6 

12- 

-2=6 

18- 

-  3=  6 

24- 

-4—6 

7- 

-1=7 

14- 

-  2=  7 

21- 

-  3=  7 

28- 

-  4—  7 

8- 

-1=8 

16- 

-  2=  8 

24- 

-  3=  8 

32- 

-  4=  8 

9- 

-1=9 

18- 

-2=9 

27- 

-  3=  9 

36- 

-  4=  9 

10- 

-1  =  10 

20- 

-2  =  10 

30- 

-3  =  10 

40- 

-  4  =  10 

11- 

-  1  =  11 

22- 

-  2  =  11 

33- 

-  3  =  11 

44- 

-  4  =  11 

12- 

-  1  =  12 

24- 

-2  =  12 

36- 

-3  =  12 

48- 

-  4  =  12 

Define  division.       Dividend.      Divisor.      Quotient.      Remainder. 
What  is  complete  division  ?    What  is  the  sign  of  division. 


DIVISION. 


55 


5- 

-  5=  1 

6- 

-  6=  1 

7- 

-  7=  1 

8- 

-  8=  1 

10- 

-  5—  2 

12- 

-6=2 

14- 

-7=2 

16- 

-  8=  2 

15- 

-  5—  3 

18- 

-  6=  3 

21- 

-7=3 

24- 

-8=3 

20- 

-  5=  4 

24- 

-6=4 

28- 

-7=4 

32- 

-  8=  4 

25- 

—  Q  mz  o 

30- 

-  6=  5 

35- 

-7=5 

40- 

-8=5 

30- 

-  5=  6 

36- 

-6=6 

42- 

-7=6 

48- 

-  8=  6 

35  — 

-  5=  7 

42- 

-6=7 

49- 

-7=7 

56- 

-8=7 

40- 

-  5    8 

48- 

-  6=  8 

56- 

-7=8 

64- 

-  8=  8 

45- 

-  5=  9 

54- 

-6=9 

63- 

-7=9 

72- 

-  8=  9 

50- 

-5  =  10 

60- 

-6  =  10 

70- 

-  7  =  10 

80- 

-  8  =  10 

55- 

-  5  =  11 

66- 

-  6  =  11 

77- 

-  7  =  11 

88- 

-  8  =  11 

60- 

-5  =  12 

72- 

-  6  =  12 

84- 

-7  =  12 

96- 

-8  =  12 

9- 

-  9=  1 

10- 

-10=  1 

11- 

-11=  1 

12- 

-12=  1 

18- 

-  9=  2 

20- 

-10=  2 

22- 

-11=  2 

24- 

-12=  2 

27- 

-9=3 

30- 

-10=  3 

33- 

-11=  3 

36- 

-12=  3 

36- 

-9=4 

40- 

-10=  4 

44- 

-11=  4 

48- 

-12=  4 

45- 

-  9=  5 

50- 

-10=  5 

55- 

-11=  5 

60- 

-12=  5 

54- 

-  9=  6 

60- 

-10=  6 

66- 

-11=  6 

72- 

-12=  6 

63- 

-  9=  7 

70- 

-10=  7 

77- 

-11=  7 

84- 

-12=  7 

72- 

-  9=  8 

80- 

-10=  8 

88- 

-11=  8 

96- 

-12=  8 

81- 

_  9=  9 

90- 

-10=  9 

99- 

-11=  9 

108- 

-12=  9 

90- 

-9  =  10 

100- 

-10  =  10 

110- 

-11  =  10 

120- 

-12=10 

99- 

-  9  =  11 

110- 

-10  =  11 

121- 

-11  =  11 

132- 

-12=11 

108- 

-  9  =  12 

120- 

-10  =  12 

132- 

-11  =  12 

144- 

-12=12 

CASE   I. 


77.   When  the  divisor  consists  of  one  figure. 
1.   How  many  times  is  4  contained  in  848  ? 


OPERATION. 


ANALYSIS.  After  writing  the  divisor 
on  the  left  of  the  dividend,  with  a  line 
between  them,  we  begin  at  the  left  hand 
and  say :  4  is  contained  in  8  hundreds, 
2  hundreds  times,  and  write  2  in  hun- 
dreds' place  in  the  quotient;  then  4  is 
contained  in  4  tens  1  ten  times,  and  write  the  1  in  tens'  place  in  the 
quotient ;  then  4  is  contained  in  8  units  2  units  times ;  and  writing  thH 
2  in  units'  place  in  the  quotient,  we  have  the  entire  quotient,  212. 


Divisor, 
Quotient, 


Dividend, 
4)848 

212 


Case  I  is  what  ?     Give  first  explanation. 


56  SIMPLE  NUMBERS. 

2.  How  many  times  is  4  contained  in  2884  ? 

OPERATION.        ANALYSIS.     As  we  cannot  divide  2  thousands  by 
4^2884          ^»  we  take  the  2  thousands  and  the  8  hundreds  to- 
gether, and  say,  4  is  contained  in  28  hundreds  7  hun- 
721          dreds  times,  which  we  write  in  hundreds'  place  in 
the  quotient ;  then  4  is  contained  in  8  tens  2  tens 
times,  which  we  write  in  tens'  place  hi  the  quotient ;  and  4  is  con- 
tained in  4  units  1  unit,  time,  which  we  write  in  units'  place  in  the 
quotient,  and  we  have  the  entire  quotient,  721. 

3.  How  many  times  is  6  contained  in  1824  ? 

OPERATION.        ANALYSIS.     Beginning  as  in  the  last  example,  we 

6)1824          say>  6  is  contained  in  18  hundreds  3  hundreds  times, 

which  we  write  in  hundreds'  place  in  the  quotient ; 

then  6  is  contained  in  2  tens  no  times,  and  we  write 

a  cipher  in  tens'  place  in  the  quotient ;  and  taking  the  2  tens  and  4 

units  together,  6  is  contained  in  24  units  4  units  times,  which  we 

write  in  units'  place  in  the  quotient,  and  we  have  304  for  the  entire 

quotient. 

4.  How  many  times  is  4  contained  in  943  ? 

OPERATION.  ANALYSIS.     Here  4  is  contained  in  9 

4)943  hundreds  2  hundreds  times,  and  1  hundred 

over,  which,  united  to  the  4  tens,  makes 

235  ...  3  Rem.        14  tens ;  4  in  14  tens,  3  tens  times  and  2 

tens  over,  which,  united  to   the  3  units, 

make  23  units ;  4  in  23  units  5  units  times  and  3  units  over.  The 
3  which  is  left  after  performing  the  division,  should  be  divided  by  4 ; 
but  the  method  of  doing  it  cannot  be  explained  until  we  reach 
Fractions ;  so  we  merely  indicate  the  division  by  placing  the  divisor 
under  the  dividend,  thus,  f .  The  entire  quotient  is  written  235|, 
which  may  be  read,  two  hundred  thirty-five  and  three  divided  by 
four,  or,  two  hundred  thirty-five  and  a  remainder  of  three. 

From  the  foregoing  examples  and  illustrations,  we  deduce 
the  following 

RULE.    I.    Write  the  divisor  at  the  left  of  the  dividend,  with 
a  line  between  them. 

Second.     Third.     Rule,  first  step? 


DIVISION. 


^e  divi- 


II.  Beginning  at  the  left  hand,  divi( 
dividend  by  the  divisor,  and  write  the 
dend. 

III.  If  there  be  a  remainder  after  dividing  any  figure,  re- 
gard it  as  prefixed  to  the  figure  of  the  next  lower  order  in  the 
dividend,  and  divide  as  before. 

IV.  Should  any  figure  or  part  of  the  dividend  be  less  than 
the  divisor,  write  a  cipher  in  the  quotient,  and  prefix  the  num- 
ber to  the  figure  of  the  next  lower  order  in  the  dividend,  and 
divide  as  before. 

V.  If  there  be  a  remainder  after  dividing  the  last  figure, 
place  it  over  the  divisor  at  the  right  hand  of  the  quotient. 

PROOF.  Multiply  the  divisor  and  quotient  together,  and  to 
the  product  add  the  remainder,  if  any ;  if  the  result  be  equal 
to  the  dividend,  the  work  is  correct. 

NOTES.  1 .  This  method  of  proof  depends  on  the  fact  that  division  is 
the  reverse  of  multiplication.  The  dividend  answers  to  the  product,  the 
divisor  to  one  of  the  factors,  and  the  quotient  to  the  othe)\ 

2.  In  multiplication  the  two  factors  are  given,  to  find  the  product : 
in  division,  the  product  and  one  of  the  factors  are  given  to  find  the 
other  factor. 


llTT 


EXAMPLES   FOR   PRACTICE. 

1.    Divide  7824  by  6. 

OPERATION. 
Divisor.      6)7824       Dividend. 

1304      Quotient. 


(2.) 

4)65432 


(3.) 
5)89135 


PROOF. 

1304       Quotient. 
6       Divisor. 


7824       Dividend. 


6)178932 


(5.) 

7)4708935 


(6.) 
8)1462376 


(7.) 
9)7468542 


Second  step?     Third?     Fourth?    Fifth?     Proof?    How  does  divis- 
ion differ  from  multiplication  ? 


58 


SIMPLE  NUMBERS. 


8.  Divide  3102455  by  5. 

9.  Divide  1762891  by  4. 

10.  Divide  546215747  by  11, 

11.  Divide  30179624  by  12. 

12.  Divide  9254671  by  9. 

13.  Divide  7341563  by  7. 


Quotients. 


Quotients. 
620491. 
440722|. 
49655977. 
2514968^. 
1028296£. 
Hem. 


14.  Divide  3179632  by  5. 

15.  Divide  19038716  by  8. 

16.  Divide  84201763  by  9. 

17.  Divide  2947691  by  12. 

18.  Divide  42084796  by  6. 

Sums  of  quotients  and  remainders,    20680083.  28. 

19.  Divide   47645  dollars   equally   among  5   men;   how 
much  will  each  receive  ?  Am.   9529  dollars. 

20.  In  one  week  are  7  days;  how  many  weeks  in  17675 
days?  Ans.   2525  weeks. 

21.  How  many  barrels  of  flour,  at  6  dollars  a  barrel,  can  be 
bought  for  6756  dollars?  Ans.    1126  barrels. 

22.  Twelve  things   make  a  dozen ;   how  many  dozen   in 
46216464?  Ans.   3851372  dozen. 

23.  How  many  barrels  of  flour  can  be  made  from  347560 
bushels  of  wheat,  if  it  take  5  bushels  to  make  one  barrel  ? 

Ans.    69512  barrels. 

24.  If  there  be  3240622  acres  of  land  in  11  townships, 
how  many  acres  in  each  township  ? 

25.  A  gentleman  left  his  estate,  worth  38470  dollars,  to  be 
shared  equally  by  his  wife  and  4  children ;  how  much  did 
each  receive  ?  Ans.   7694  dollars. 

CASE    II. 

78.  When  the  divisor  consists  of  two  or  more  figures. 

NOTE.   To  illustrate  more  clearly  the  method  of  operation,  we  will 
first  take  an  example  usually  performed  by  Short  Division. 


Case  II  is  what  ? 


DIVISION.  59 

1.  How  many  times  is  8  contained  in  2528  ? 
OPERATION.  ANALYSIS.    As  8  is  not  contained  in  2  thou- 

8  }  2528  (  316        sands,  we  take  2  and  5  as  one  number,  and  con- 
9  A  sider  hoAv  many  times  8  is  contained  in  this 

partial  dividend,  25  hundreds,  and  find  that  it 
12  is  contained  3  hundreds  times,  and  a  remainder. 

8  To  find  this  remainder,  we  multiply  the  divisor, 

8,  by  the  quotient  figure,  3  hundreds,  and  sub- 
tract the  product,  24  hundreds,  from  the  par- 
tial dividend,  25  hundreds,  and  there  remains 
1  hundred.     To  this  remainder  we  bring  down 
the  2  tens  of  the  dividend,  and  consider  the  12  tens  a  second  partial 
dividend.     Then,  8  is  contained  in  12  tens  1  ten  time  and  a  remain- 
der ;  8  multiplied  by  1  ten  produces  8  tens,  which,  subtracted  from, 
12  tens,  leave  4  tens.     To  this  remainder  we  bring  down  the  8  units, 
and  consider  the  48  units  the  third  partial  dividend.     Then, 8  is  con- 
tained in  48  units  6  units  times.     Multiplying  and  subtracting  as 
before,  we  find  that  nothing  remains,  and  we  have  for  the  entire 
quotient,  316. 

2.  How  many  times  is  23  contained  in  4807  ? 

OPERATION.  ANALYSIS.     We  first  find  how 

Divisor.  Divid'd.  Quotient.  many  times  23  is  contained  in  48, 

23  )  4807  (  209  the  first  partial  dividend,  and  place 

46  the  result  in  the  quotient  on  the 

~TTT  right  of  the   dividend.     We  then 

multiply  the   divisor,    23,   by  the 

quotient  figure,  2,  and  subtract  the 

product,  46,  from  the  part  of  the 

dividend  used,  and  to  the  remainder  bring  down  the  next  figure  of 
the  dividend,  which  is  0,  making  20,  for  the  second  partial  dividend. 
Then,  since  23  is  contained  in  20  no  times,  we  place  a  cipher  in  the 
quotient,  and  bring  down  the  next  figure  of  the  dividend,  making  a 
third  partial  dividend,  207  ;  23  is  contained  in  207,  9  times ;  multi- 
plying and  subtracting  as  before,  nothing  remains,  and  we  have  for 
the  entire  quotient,  209. 

NOTES.  1 .  When  the  process  of  dividing  is  performed  mentally,  and 
the  results  only  are  written,  as  in  Case  I,  the  operation  is  termed  Short 
Division. 

2.  When  the  whole  process  of  division  is  written,  the  operation  is 
termed  Long  Division. 

Give  first  explanation.  Second.  What  is  long  division  ?  What  is 
short  division  r  When  is  each  used  r 


60  SIMPLE   NUMBERS. 

3.  Short  Division  is  generally  used  when  the  divisor  is  a  number 
that  will  allow  the  process  of  dividing  to  be  performed  mentally. 

From  the  preceding  illustrations  we  derive  the  following 
general 

RULE.  I.  Write  the  divisor  at  the  left  of  the  dividend,  as 
in  short  division. 

II.  Divide  the  least  number  of  the  left  hand  figures  in  the 
dividend  that  will  contain  the  divisor  one  or  more  times ',  and 
place  the  quotient  at  the  right  of  the  dividend,  with  a  tine  be- 
tween them. 

III.  Multiply  the  divisor  by  this  quotient  figure,  subtract 
the  product  from  the  partial  dividend  used,  and  to  the  remain- 
der bring  down  the  next  figure  of  the  dividend. 

IV.  Divide  as  before,  until  all  the  figures  of  the  dividend 
have  been  brought  down  and  divided. 

V.  If  any  partial  dividend  will  not  contain  the  divisor, 
place  a  cipher  in  the  quotient,  and  bring  down  the  next  figure 
of  the  dividend,  and  divide  as  before. 

VI.  If  there  be  a  remainder  after  dividing  all  the  figures  of 
the  dividend,  it  must  be  written  in  the  quotient,  with  the  divi- 
sor underneath. 

NOTES.  1.  If  any  remainder  be  equal  to,  or  greater  than  the  divisor, 
the  quotient  figure  is  too  small,  and  must  be  increased. 

2.  If  the  product  of  the  divisor  by  the  quotient  figure  be  greater 
than  the  partial  dividend,  the  quotient  figure  is  too  large,  and  must  be 
diminished. 

79.  PROOF.     1.  The  same  as  in  short  division.     Or, 

2.  Subtract  the  remainder,  if  any,  from  the  dividend,  and 
divide  the  difference  by  the  quotient ;  if  the  result  be  the  same 
as  the  given  divisor,  the  work  is  correct. 

80.  The  operations  in  long  division  consist  of  five  prin- 
cipal steps,  viz. :  — 

1st.    Write  down  the  numbers. 

Rule,  first  step  ?  Second  ?  Third  ?  Fourth  ?  Fifth  ?  Sixth  ?  First 
direction  ?  Second  ?  Proof?  Recapitulate  the  steps  in  their  order. 


DIVISION.  61 

2d.  Find  how  many  times. 

3d.  Multiply. 

4th.  Subtract. 

5th.  Bring  down  another  figure. 

EXAMPLES    FOR   PRACTICE. 

3.  Find  how  many  times  36  is  contained  in  11798. 

OPERATION.  PROOF  BY  MULTIPLICATION. 

Dividend. 

Divisor.      36)  1179#'(327  Quotient.  327       Quotient. 

108J'  36        Divisor. 

9S17 

72 

2^ 

252  26        Remainder. 

26  Remainder.  11798        Dividend. 

4.  Find  how  many  times  82  is  contained  in  89634. 

OPERATION.  PROOF  BY  DIVISION. 

82  )  896£4-(  1093  89634       Dividend. 

82  8          Remainder. 

76^1  Quotient.        1093  )  89626  (  82       Divisor. 

738  8744 

~254  2186 

246  2186 


5.  Find  how  many  times  154  is  contained  in  32740. 

6.  Divide  32572  by  34.  Ans.  958. 

7.  Divide  1554768  by  216.  Ans.  7198. 

8.  Divide  5497800  by  175.  Ans.  31416. 

9.  Divide  3931476  by  556.  Ans.  7071. 
10.  Divide  10983588  by  132.  Ans.  83209. 

F 


62  SIMPLE  NUMBERS. 

11.  Divide  73484248  by  19.  Am.  3867592. 

12.  Divide  8121918  by  21.  Am.  386758. 

13.  Divide  10557312  by  16.  Ans.   659832. 

14.  Divide  93840  by  63.  Rem.   33. 

15.  Divide  352417  by  29.  Rem.    9. 

16.  Divide  51846734  by  102.  Hem.   32. 

17.  Divide  1457924651  by  1204.  Rem.  1051. 

18.  Divide  729386  by  731.  Rem.  579. 

19.  Divide  4843167  by  3605.  Rem.  1652. 

20.  Divide  49816657  by  9101.  Rem.  6884. 

21.  Divide  75867308  by  10115.  Rem.  4808. 

Quotients.    Rem. 

22.  Divide  28101418481  by  1107,         25385201.         974. 

23.  Divide  65358547823  by  2789.        23434402.         645. 

24.  Divide  102030405060  by  123456.      826451.     70404. 

25.  Divide  48659910  by  54001.  901.       5009. 

26.  Divide  2331883961  by  6739549.  346.  7. 

27.  A  railroad  cost  one  million  eight  hundred  fifty  thousand 
four  hundred  dollars,  and  was  divided  into  eighteen  thousand 
five  hundred  and  four  shares ;  what  was  the  value  of  each 
share?  Ans.    100  dollars. 

28.  If  a  tax  of  seventy-two  million  three  hundred  twenty 
thousand  sixty  dollars  be  equally   assessed  on  ten  thousand 
seven  hundred  thirty-five  towns,  what  amount  of  tax  must 
each  town  pay  ?  Ans.    6736TVV^  dollars. 

29.  In  1850  there  were  in  the  United  States  213  college 
libraries,  containing   942321    volumes  ;    what  would  be  the 
average  number  of  volumes  to  each  library  ? 

Ans.    442 4?f  ^  vols. 

30.  The  number  of  post  offices  in  the  United  States  in 
1853  was  22320,  and  the  entire  revenue  of  the  post  office 
department   was   5937120   dollars;    what  was   the   average 
revenue  of  each  office?  Ans.  266  dollars. 


DIVISION.  63 

CONTRACTIONS. 

CASE    I. 

81.  When  the  divisor  is  a  composite  number. 

1.  If  3270  dollars  be  divided  equally  among  30  men,  how 
many  dollars  will  each  receive  ? 

OPERATION.  ANALYSIS.      If   3270  dollars   be    divided 

5)3270  equally  among  30  men,  each  man  will  receive 

as  many  dollars  as  30  is  contained  times  in 

6)654  3270  dollars.     30  may  be  resolved  into  the 

109  Ans.         factors  5  and  6  ;  and  we  may  suppose  the  30 

men  divided  into  5  groups  of  6  men   each ; 

dividing  the  3270  dollars  by  5,  the  number  of  groups,  we  have 

654,  the  number  of  dollars  to  be  given  to  each  group ;  and  dividing 

the  654  dollars  by  6,  the  number  of  men  in  each  group,  we  have 

109,  the  number  of  dollars  that  each  man  will  receive.     Hence, 

RULE.  Divide  the  dividend  by  one  of  the  factors,  and  the 
quotient  thus  obtained  by  another,  and  so  on  if  there  be  more 
than  two  factors,  until  every  factor  has  been  made  a  divisor. 
The  last  quotient  will  be  the  quotient  required. 

EXAMPLES    FOR   PRACTICE. 

2.  Divide  3690  by  15  =  3  X  5.  Ans.  246. 

3.  Divide  3528  by  24  =  4  X  6.  Ans.  147. 

4.  Divide  7280  by  35  =  5  X  7.  Ans.  208. 

5.  Divide  6228  by  36  =  6  X  6.  Ans.  173. 

6.  Divide  33642  by  27  =  3  X  9.  Ans.  1246. 

7.  Divide  153160  by  56  —  7  X  8.  Ans.  2735. 

8.  Divide  15625  by  125  =  5  X  5  X  5.       Ans.  125. 

82.  To  find  the  true  remainder. 

1.  Divide  1143  by  64,  using  the  factors  2,  8,  and  4,  and  find 
the  true  remainder. 

What  are  contractions  ?     Case  I  is  what  ?    Give  explanation.     Rule. 


64  SIMPLE  NUMBERS. 

OPERATION.  ANALYSIS.    Divid- 

2)1143  ing   1143    by   2,   we 

Q7777  -,  have    a    quotient    of 

8)571 1  rem.  .,...        ,^         .   . 

L 571,  and  a  remainder 

4)71- 3X2=6    "  of  1  undivided,  which, 

17        q  v  ft  V  9  —  Aft      a  being    a    Part    of   the 

-  ™  given  dividend,  must 

55  true  rem.      also  be  a  part  of  the 

true  remainder.     The 

571   being  a  quotient   arising  from  dividing  by  2,  its  units   are 

2  times  as  great  in  value  as  the  units  of  the  given  dividend,  1 143. 
Dividing  the  571  by  8,  we  have  a  quotient  of  71,  and  a  remainder 
of  3  undivided.     As  this  3  is  a  part  of  the  571,  it  must  be  multiplied 
by  2  to  change  it  to  the  same  kind  of  units  as  the  1.     This  makes  a 
true  remainder  of  6  arising  from  dividing  by  8.     Dividing  the  7 1  by 
4,  we  have  a  quotient  of  17,  and  a  remainder  of  3  undivided.     This 

3  is  a  part  of  the  71,  the  units  of  which  are  8  times  as  great  in  value 
as  those  of  the  571,  and  the  units  of  the  571  are  2  times  as  great 
in  value  as  those  of  the  given  dividend,  1143 ;  therefore,  to  change 
this  last  remainder,  3,  to  units  of  the  same  value  as  the  dividend, 
we  multiply  it  by  8  and  2,  and  obtain  a  true  remainder  of  48  arising 
from  dividing  by  4.     Adding  the  three  partial  remainders,  we  obtain 
55,  the  true  remainder.     Hence, 

RULE.  I.  Multiply  each  partial  remainder  by  all  the  pre- 
ceding divisors. 

II.  Add  the  several  products,  and  the  sum  will  be  the  true 
remainder. 

EXAMPLES    FOR   PRACTICE. 


Hem, 

2. 

Divide 

34712 

by 

42 

—  6 

X 

7. 

20. 

3. 

Divide 

401376 

by 

64 

—  8 

X 

8 

. 

32. 

4. 

Divide 

139074 

by 

72 

=  3 

X 

4 

X 

6. 

42. 

5. 

Divide 

9078126 

by 

90 

=  3 

X 

5 

X 

6. 

6. 

6. 

Divide 

18730627 

by 

120 

=  4 

X 

5 

X 

6. 

67. 

7. 

Divide 

7360479 

by 

96 

n 

X 

6 

X 

8. 

63. 

8. 

Divide 

24726300 

by 

70 

—  2 

X 

5 

X 

7. 

60. 

9. 

Divide 

5010207 

by 

84 

rr 

X 

2 

X 

6. 

15. 

Explain  the  process  of  finding  the  true  remainder  when  dividing  by 
the  factors  of  a  composite  number. 


DIVISION.  65 

CASE   II. 

83.  When  the  divisor  is  10,  100,  1000,  &c. 

1.  Divide  374  acres  of  land  equally  among  10  men ;  how 
many  acres  will  each  have  ? 

OPERATION.  ANALYSIS.     Since  we  have  shown, 

110)3714  ^at  ^°  remove  a  figure  one  place 

toward  the  left  by  annexing  a  cipher 

Quotient.       37 4  Rem.        increases  its  value  tenfold,  or  multi- 

or,  37 T4<y  acres.  plies  it  by  10,  (68,)  so,  on  the  con- 

trary, by  cutting  off  or  taking  away 

the  right  hand  figure  of  a  number,  each  of  the  other  figures  is  removed 
one  place  toward  the  right,  and,  consequently,  the  value  of  each  is 
diminished  tenfold,  or  divided  by  10,  (32.) 

For  similar  reasons,  if  we  cut  off  two  figures,  we  divide  by 
100,  if  three,  we  divide  by  1000,  and  so  on.  Hence  the 

RULE.  From  the  right  hand  of  the  dividend  cut  off  as 
many  figures  as  there  are  ciphers  in  the  divisor.  Under  the 
figures  so  cut  off,  place  the  divisor,  and  the  whole  will  form  the 
quotient. 

EXAMPLES   FOR  PRACTICE. 

2.  Divide  4760  by  10. 

3.  Divide  362078          by  100. 

4.  Divide  1306321        by  1000. 

5.  Divide  9760347        by  10000. 

6.  Divide  2037160310  by  100000. 

CASE    III. 

84.  When  there  are  ciphers  on  the  right  hand  of 
the  divisor. 

1.   Divide  437661  by  800. 

OPERATION.  ANALYSIS.    In  this  example  we 

8 1 00) 437 6 1 61  resolve  800  into  the  factors  8  and 

100,  and  divide  first  by  100,  by  cut- 

547  -  -  -  61  Rem.  ting  Ofl-  two  right  hand  figures  of  the 

Case  II  is  what?  Give  explanation.  Rule.  Case  in  is  what£ 
Give  explanation. 


66  SIMPLE  NUMBERS. 

dividend,  (83,)  and  we  have  a  quotient  of  4376,  and  a  remainder  of 
61.  We  next  divide  by  8,  and  obtain  547  for  a  quotient;  and  the 
entire  quotient  is 


2.   Divide  34716  by  900. 

OPERATION.  ANALYSIS.      Dividing 

9|00)347|16  as  in  the  last  example,  we 

have  a  quotient  of  38,  and 
38   Quotient.          5,  2d  rem.  ^  remainders>   16    and 

5  X  100  +  16  =  516,  true  rem.  5,       Multiplying    5,    the 

38££§,  Ans.  last  remainder,  by   100, 

the  preceding  divisor,  and 

adding  16,  the  first  remainder,  (82,)  we  have  516  for  the  true  re- 
mainder. But  this  remainder  consists  of  the  last  remainder,  5,  pre- 
fixed to  the  figures  16,  cut  off  from  the  dividend.  Hence, 

85.  When  there  is  a  remainder  after  dividing  by  the  sig- 
nificant figures,  it  must  be  prefixed  to  the  figures  cut  off  from 
the  dividend  to  give  the  true  remainder  ;  if  there  be  no  other 
remainder,  the  figures  cut  off  from  the  dividend  will  be  the 
true  remainder. 

EXAMPLES    FOR   PRACTICE. 

Quotients.       Bern. 

3.  Divide  34716  by  900.  38           516 

4.  Divide  1047634        .  by  2400.  436         1234 

5.  Divide  47321046  by  45000.  1051  26 

6.  Divide  2037903176  by  140000.  63176 

7.  Divide  976031425  'by  92000.  3425 

8.  Divide  80013176321  by  700000.  376321 

9.  Divide  19070367428  by  4160000.  4584     927428 

10.  Divide  379025644319  by  554000000.         535644319 

11.  The  circumference  of  the  earth  at  the  equator  is  24898 
miles.    How  many  hours  would  a  train  of  cars  require  to  travel 
that  distance,  going  at  the  rate  of  50  miles  an  hour  ? 

Ans.   497  f*. 

12.  The  sum  of  350000  dollars  is  paid  to  an  army  of  14000 
men  ;  what  does  each  man  receive  ?  Ans.    25  dollars. 

How  is  the  true  remainder  found  ? 


PROMISCUOUS   EXAMPLES.  67 


EXAMPLES    IN    THE    PRECEDING    RULES. 

1.  George  Washington  was  born  in  1732,  and  lived  67 
years  ;  in  what  year  did  he  die  ?  Ans.   in  1799. 

2.  How  many  dollars  a  day  must  a  man  spend,  to  use  an 
income  of  1095  dollars  a  year  ?  Ans.   3  dollars. 

3.  If  I  give  141  dollars  for  a  piece  of  cloth  containing  47 
yards,  for  how  much  must  I  sell  it  in  order  to  gain  one  dollar 
a  yard  ?  Ans.    188  dollars. 

4.  A  speculator  who  owned  500  acres,  17  acres,  98  acres, 
and  121  acres  of  land,  sold  325  acres  ;  how  many  acres  had 
he  left?  Ans.   411  acres. 

5.  A  dealer  sold  a  cargo  of  salt  for  2300  dollars,  and  gained 
625  dollars  ;  what  did  the  cargo  cost  him  ? 

Ans.   1675  dollars. 

6.  If  a  man  earn  60  dollars  a  month,  and  spend  45  dol- 
lars in  the  same  time,  how  long  will  it  take  him  to  save  900 
dollars  from  his  earnings  ? 

7.  If  9  persons  use  a  barrel  of  flour  in  87  days,  how  many 
days  will  a  barrel  last  1  person  at  the  same  rate  ? 

Ans.   783  days. 

8.  The  first  of  three  numbers  is  4,  the  second  is  8  times 
the  first,  and  the  third  is  9  times  the  second ;  what  is  their 
sum?  Ans.   324. 

9.  If  2,  2,  and  7  are  three  factors  of  364,  what  is  the 
other  factor?  Ans.   13. 

10.  A  man  has  3  farms ;  the  first  contains  78  acres,  the 
second  104  acres,  and  the  third  as  many  acres  as  both  the 
others  ;  how  many  acres  in  the  3  farms  ? 

11.  If  the  expenses  of  a  boy  at  school  are  90  dollars  for 
board,  30  dollars  for  clothes,  12  dollars  for  tuition,  5  dollars 
for  books,  and  7  dollars  for  pocket  money,  what  would  be  the 
expenses  of  27  boys  at  the  same  rate  ?     Ans.  3888  dollars. 

12.  Four  children  inherited  2250  dollars  each;  but  one 
dying,  the  remaining  three  inherited  the  whole ;  what  was  the 
share  of  each?  Ans.   3000  dollars. 


68  SIMPLE  NUMBERS. 

13.  Two  men  travel  in  opposite  directions,  one  at  the  rate 
of  35  miles  a  day,  and  the  other  at  the  rate  of  40  miles  a  day  ; 
how  far  apart  are  they  at  the  end  of  6  days  ? 

14.  Two  men  travel  in  the  same  direction,  one  at  the  rate 
of  35  miles  a  day,  and  the  other  at  the  rate  of  40  miles  a 
day ;  how  far  apart  are  they  at  the  end  of  6  days  ? 

15.  A  man  was  45  years  old,  and  he  had  been  married  19 
years ;  how  old  was  he  when  married?         Ans.   26  years. 

16.  Upon  how  many  acres  of  ground  can  the  entire  popu- 
lation of  the  globe  stand,  supposing  that  25000  persons  can 
stand  upon  one  acre,  and  that  the  population  is  1000000000  ? 

Ans.    40000  acres. 

17.  Add  384,  1562,  25,  and-  946 ;  subtract  2723  from  the 
sum ;  divide  the  remainder  by  97  ;  and  multiply  the  quotient 
by  142  ;  what  is  the  result  ?  Ans.   284. 

18.  How  many  steps  of  3  feet  each  would  a  man  take  in 
•walking  a  mile,  or  5280  feet?  Ans.    1760  steps. 

19.  A  man  purchased  a  house  for  2375  dollars,  and  ex- 
pended 340  dollars  in  repairs ;  he  then  sold  it  for  railroad 
stock  worth  867  dollars,  and  235  acres  of  western  land  val- 
ued at  8  dollars  an  acre  ;  how  much  did  he  gain  by  the  trade  ? 

Ans.   32  dollars. 

20.  The  salary  of  a  clergyman  is  800  dollars  a  year,  and 
his  yearly  expenses  are  450  dollars;  if  he  be  worth  1350  dol- 
lars now,  in  how  many  years  will  he  be  worth  4500  dollars  ? 

Ans.   9  years. 

21.  How  many  bushels  of  oats  at  40  cents  a  bushel,  must 
be  given  for  1600  bushels  of  wheat  at  75  cents  a  bushel  ? 

Ans.   3000  bushels. 

22.  Bought  325  loads  of  wheat,  each  load  containing  50 
bushels,  at  2  dollars  a  bushel ;  what  did  the  wheat  cost  ? 

23.  If  you  deposit  225  cents  each  week  in  a  savings  bank, 
and  take  out  75  cents  a  week,  how  many  cents  will  you  have 
left  at  the  end  of  the  year  ?  Ans.   7800  cents. 

24.  The  product  of  two  numbers  is  31383450,  and  one  of 
the  numbers  is  4050  ;  what  is  the  other  number  ? 


PROMISCUOUS   EXAMPLES.  69 

25.  The  Illinois  Central  Railroad  is  700  miles  long,  and 
cost  31647000  dollars ;  what  did  it  cost  per  mile  ? 

Ans.    45210  dollars. 

26.  What  number  is  that,  which  being  divided  by  7,  the 
quotient  multiplied  by  3,  the  product  divided  by  5,  and  this 
quotient  increased  by  40,  the  sum  will  be  100  ?    Ans.   700. 

27.  How  many  cows   at  27  dollars  apiece,  must  be  given 
for  54  tons  of  hay  at  17  dollars  a  ton  ? 

28.  A  mechanic  receives  56  dollars  for  26  days'  work,  and 
spends  2  dollars  a  day  for  the  whole  time  ;  how  many  dollars 
has  he  left  ?  Ans.  4  dollars. 

29.  If  7  men  can  build  a  house  in  98  days,  how  long  would 
it  take  one  man  to  build  it  ?  Ans.    686  days. 

30.  The   number   of  school  houses  in  the  State  of  New 
York,  in  1855,  was  11,137  ;  suppose  their  cash  value  to  have 
been  5,301,212  dollars,  what  would  be  the  average  value? 

Ans.   476  dollars. 

31.  A  thief,  traveling  5  miles  an  hour,  has  48  miles  the 
start  of  an  officer  who  is  pursuing  him  at  the  rate  of  1 1  miles 
an  hour  ;  in  how  many  hours  will  the  officer  overtake  him  ? 

Ans.    8  hours. 

32.  The  average  beat  of  the  pulse  of  a  man  at  middle  age 
is  about  4500  times  in  an  hour  ;  how  many  times  does  it  beat 
in  24  hours  ?  Ans.    108000  times. 

33.  How  many  years  from  the  discovery  of  America,  in 
1492,  to  the  year  1900  ? 

34.  According   to   the  census,   Maine   has   31766   square 
miles;  New  Hampshire,  9280;  Vermont,  10212;  Massachu- 
setts, 7800;   Rhode  Island,   1306;    Connecticut,  4674;   and 
New  York,  47000 ;    how  many  more  square  miles  has  all 
New  England  than  New  York? 

35.  What  is  the  remainder  after  dividing  62530000   by 
87900?  Ans.  33100. 

36.  A  pound  of  cotton  has  been  spun  into  a  thread  8  miles 
in  length  ;  allowing  235  pounds  for  waste,  how  many  pounds 
will  it  take  to  spin  a  thread  to  reach  round  the  earth,  suppos- 
ing the  distance  to  be  25000  miles?        Ans.   3360  pounds. 


70  SIMPLE   NUMBERS. 

37.  John  has  8546  dollars,  which  is  342  dollars  less  than 
4  times  as  much  as    Charles   has  ;   how  many  dollars  has 
Charles  ?  Am.    2222  dollars. 

38.  The  quotient  of  one  number  divided  by  another  is  37, 
the  divisor  245,  and  the  remainder  230;  what  is  the  divi- 
dend? Ans.   9295. 

39.  What    number    multiplied    by    72084   will    produce 
5190048?  Ans.   72. 

40.  There  are  two  numbers,  the  greater  of  which  is  73 
times  109,  and  their  difference  is  17  times  28 ;  what  is  the  less 
number?  Ans.   7481. 

41.  The  sum  of  two  numbers  is  360,  and  the  less  is  114 ; 
what  is  the  product  of  the  two  numbers  ?          Ans.   28044. 

42.  What  number  added  to  2473248  makes  2568754? 

Ans.    95506. 

43.  A  farmer  sold  35  bushels  of  wheat  at  2  dollars  a  bush- 
el, and  18  cords  of  wood  at  3  dollars  a  cord;  he  received  9 
yards  of  cloth  at  4  dollars  a  yard,  and  the  balance  in  money ; 
how  many  dollars  did  he  receive  ?  Ans.   88  dollars. 

44.  A  cistern  whose  capacity  is  840  gallons  has  two  pipes ; 
through  one  pipe  60  gallons  run  into  it  in  an  hour,  and  through 
the  other  39  gallons  run  out  in  the  same  time ;  in  how  many 
hours  will  the  cistern  be  filled  ?  Ans.   40  hours. 

45.  Two  men  start  from  the  same  place,  and  travel  in  the 
same  direction  ;  one  travels  32  miles  a  day,  and  the  other  46; 
how  far  will  they  be  apart  at  the  end  of  15  days  ? 

Ans.    210  miles. 

46.  Henry  resides  1  mile  from  school,  and  George  2  miles  ; 
if  they  each  attend  school  295  days,  how  many  more  miles 
will  George  travel  in  that  time  than  Henry  ?  Ans  590  miles. 

47.  If  the  remainder  be  3252,  and  the  minuend  4248,  what 
is  the  subtrahend  ?  Ans.    996. 

48.  How  many  times  will  a  coach  wheel  16  feet  in  circum- 
ference revolve  in  going  160  miles,  there  being  5280  feet  in  a 
mile  ?  Ans.   52800  times. 


PROMISCUOUS  EXAMPLES.  71 

49.  A  salt  dealer  has  3524  barrels  in  store;   how  many 
barrels  will  he  have  left  after  selling  75  barrels  to  one  man, 
149  to  another,  854  to  another,  and  287  to  another? 

Am.   2159  barrels. 

50.  How  many  yards  of  muslin  will  it  take  to  make  dresses 
for  a  family  of  7  girls,  each  girl  having  3  dresses,  and  each 
dress  averaging  9  yards?  Ans.    189  yards. 

51.  If  a  man  can  do  a  piece  of  work  in  855  days,  how 
many  men  can  do  it  in  45  days?  Ans.    19  men. 

52.  A  man,  traveling  27  miles  a  day,  starts  from  Boston 
to  meet  another  man,  traveling  45  miles  a  day,  from  New  York, 
216  miles  distant;  in  how  many  days  will  they  meet ? 

Ans.   3  days. 

53.  The  population  of  New  York  in  1855  was  3466212, 
of  which  1124211    attended  church   regularly;   how    many 
persons  in  the  state  were  not  regular  attendants  at  church  ? 

Ans.   2342001  persons. 

54.  A  farmer  receives  684  dollars  a  year  for  produce  from 
his  farm,  and  his  expenses  are  375  dollars  a  year ;  how  many 
dollars  will  he  save  in  five  years  ? 

55.  The  salt  manufacturer  at  Syracuse  pays  58  cents  for 
wood  to  boil  one  barrel  of  salt,  10  cents  for  boiling,  5  cents  to 
the  state  for  the  brine,  28  cents  for  the  packing  barrel,  and  3 
cents  for  packing  and  weighing,  and  receives  125  cents  from 
the  purchaser ;  how  many  cents  does  he  make  on  a  barrel  ? 

Ans.   21  cents. 

56.  A  company  of   15    persons  purchase  a  township   of 
western  land  for  286000  dollars,  of  which  sum  one  man  pays 
6000  dollars,  and  the  others  the  remainder,  in  equal  amounts ; 
how  much  does  each  of  the  others  pay  ?     Ans.    20000  dollars. 

57.  If  256  be  multiplied  by  25,  the  product  diminished  by 
625,  and  the  remainder  divided  by  35,  what  will  be  the  quo- 
tient? Ans.    165. 

58.  Two  men  start  from  different  places,  distant  189  miles, 
and  travel  toward  each  other ;  one  goes  4  miles,  and  the  other 
5  miles  an  hour ;  in  how  many  hours  will  they  meet  ? 


72  SIMPLE   NUMBERS. 


GENERAL  PRINCIPLES   OF  DIVISION. 

8G.  The  quotient  in  Division  depends  upon  the  relative 
values  of  the  dividend  and  divisor.  Hence  any  change  in  the 
value  of  either  dividend  or  divisor  must  produce  a  change  in 
the  value  of  the  quotient.  But  some  changes  may  be  produced 
upon  both  dividend  and  divisor,  at  the  same  time,  that 
will  not  affect  the  quotient.  The  laws  which  govern  these 
changes  are  called  General  Principles  of  Division,  which  we 
will  now  examine. 

I.  54-^-9  =  6. 

If  we  multiply  the  dividend  by  3,  we  have 
54  x  3  -^  9  =  162  -±-  9  =  18, 

and  18  equals  the  quotient,  6,  multiplied  by  3.  Hence, 
Multiplying  the  dividend  by  any  number,  multiplies  the  quotient 
by  the  same  number. 

II.  Using  the  same  example,  54  -7-  9  =  6. 
If  we  divide  the  dividend  by  3  we  have 

V-T-9  =  18-r-9=  2, 

and  2  =  the  quotient,  6,  divided  by  3.  Hence,  Dividing  the 
dividend  by  any  number^  divides  the  quotient  by  the  same 
number. 

III.  If  we  multiply  the  divisor  by  3,  we  have 

54  _L.  9  X  3  =  54  ~  27  —  2, 

and  2  =  the  quotient,  6,  divided  by  3.  Hence,  Multiplying 
the  divisor  by  any  number,  divides  the  quotient  by  the  same 
number. 

IV.  If  we  divide  the  divisor  by  3,  we  have 

5*  -T-|  =  54-^8  =  18, 


Upon  what  does  the  value  of  the  quotient  depend  ?     What  is  the 
first  general  principle  of  division?      Second?      Third?      Fourth? 


GENERAL  PRINCIPLES  OF  DIVISION.  73 

and  18  r=  the  quotient,  6,  multiplied  by  3.  Hence,  Dividing 
the  divisor  by  any  number,  multiplies  the  quotient  by  the  same 
number. 

V.  If  we  multiply  both  dividend  and  divisor  by  3,  we  have 

54  X  3  -^  9  X  3  =  162  -^-  27  —  6. 

Hence,  Multiplying  both  dividend  and  divisor  ty  ike  same  num- 
ber, does  not  alter  the  value  of  the  quotient. 

VI.  If  we  divide  both  dividend  and  divisor  by  3,  we  have 

jyt^f  =18 -f.3=6. 

Hence,  Dividing  both  dividend  and  divisor  by  the  same  num- 
ber, does  not  alter  the  value  of  the  quotient. 

87.  These  six  examples  illustrate  all  the  different  changes 
we  ever  have  occasion  to  make  upon  the  dividend  and  divisor 
in    practical   arithmetic.     The    principles    upon  which   these 
changes  are  based  may  be  stated  as  follows  : 

PRIN.  I.  Multiplying  the  dividend  multiplies  the  quotient ; 
and  dividing  the  dividend  divides  the  quotient.  (8G.  I  and  II.) 

PRIN.  II.  Multiplying  the  divisor  divides  the  quotient  ,•  and 
dividing  the  divisor  multiplies  the  quotient.  (8G.  Ill  and  IV.) 

PRIN.  III.  Multiplying  or  dividing  both  dividend  and 
divisor  by  the  same  number,  does  not  alter  the  quotient.  (8G. 
V  and  VI.) 

88.  These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

A-  change  in  the  dividend  produces  a  LIKE  change  in  the 
quotient;  but  a  change,  in  the  divisor  produces  an  OPPOSITE 
change  in  the  quotient. 

NOTE.  If  a  number  be  multiplied  and  the  product  divided  by  the 
same  number,  the  quotient  will  be  equal  to  the  number  multiplied. 
Thus,  15  X  4  =  60,  and  60  -f-  4  =  15. 

Fifth  ?  Sixth  ?  Into  how  many  general  principles  can  these  be  con- 
densed ?  What  is  the  first  ?  Second  ?  Third  ?  In  what  general  law 
are  these  embraced  ? 


74  PROPERTIES   OF   NUMBERS. 


EXACT   DIVISORS. 

80o  An  Exact  Divisor  of  a  number  is  one  that  gives 
a  whole  number  for  a  quotient. 

As  it  is  frequently  desirable  to  know  if  a  number  has  an  exact 
divisor,  we  will  present  a  few  directions  that  will  be  of  assistance, 
particularly  in  finding  exact  divisots  of  large  numbers. 

NOTE.  A  number  whose  unit  figure  is  0,  2,  4,  6,  or  8  is  called  an 
Even  Number.  And  a  number  whose  unit  figure  is  1,  3,  5,  7,  or  9,  is 
called  an  Odd  Number. 

2  is  an  exact  divisor  of  all  even  numbers. 

4  is  an  exact  divisor  when  it  will  exactly  divide  the  tens 
and  units  of  a  number.     Thus,  4  is  an  exact  divisor  of  268, 
756,  1284. 

5  is  an  exact  divisor  of  every  number  whose  unit  figure  is 
0  or  5.     Thus,  5  is  an  exact  divisor  of  20,  955,  and  2840. 

8  is  an  exact  divisor  when  it  will  exactly  divide  the  hun- 
dreds, tens,  and  units   of  a  number.     Thus,  8  is  an  exact 
divisor  of  1728,  5280,  and  213560. 

9  is  an  exact  divisor  when  it  will  exactly  divide  the  sum  of 
the  digits  of  a  number.     Thus,  in  2486790,  the  sum  of  the 
digits  2  +  4+8  +  6  +  7+9  +  0  =  36,  and  36  -J-  9  —  4. 

10  is  an  exact  divisor  when  0  occupies  units'  place. 
100  when  00  occupy  the  places  of  units  and  tens. 

1000  when  000  occupy  the  places  of  units,  tens,  and  hun- 
dreds, &c. 

A  composite  number  is  an  exact  divisor  of  any  number, 
when  all  its  factors  are  exact  divisors  of  the  same  number. 
Thus,  2,  2,  and  3  are  exact  divisors  of  1 2 ;  and  so  also  are  4 
(=2X2)  and  6  (=  2  X  3). 

An  even  number  is  not  an  exact  divisor  of  an  odd  number. 

If  an  odd  number  is  an  exact  divisor  of  an  even  number, 

What  is  an  exact  divisor  ?  "What  is  an  even  number  ?  An  odd  num- 
ber ?  When  is  2  an  exact  divisor  ?  4  ?  5  ?  9  ?  10  ?  100  ?  1000  ? 
When  is  a  composite  number  an  exact  divisor  ?  An  even  number  is 
not  an  exact  divisor  of  what  ?  An  odd  number  is  an  exact  divisor  of 
•what? 


FACTORING    NUMBERS. 


75 


twice  that  odd  number  is  also  an  exact  divisor  of  the  even 
number.  Thus,  7  is  an  exact  divisor  of  42  ;  so  also  is  7  X  2, 
or  14. 

PRIME    NUMBERS. 

9O.    A  Prime  Number  is  one  that  can  not  be  resolved 
or  separated  into  two  or  more  integral  factors. 

For  reference,  and  to  aid  in  determining  the  prime  factors 
of  composite  numbers,  we  give  the  following :  — 

TABLE  OF  PRIME  NUMBERS  FROM   1  TO   1000. 


1 

59 

139 

233 

337 

439 

557 

653 

769 

883 

2 

61 

149 

239 

347 

443 

563 

659 

773 

.887 

3 

67 

151 

241 

349 

449 

569 

661 

787 

907 

5 

71. 

157 

251 

353 

457 

571 

673 

797 

911 

7 

73 

163 

257 

359 

461 

577 

677 

809 

919 

11 

79 

167 

263 

367 

463 

587 

683 

811 

929 

13 

83 

173 

269 

373 

467 

593 

691 

821 

937 

17 

89 

179 

271 

379 

479 

599 

701 

823 

941 

19 

97 

181 

277 

383 

487 

601 

709 

827 

947 

23 

101 

191 

281 

389 

491 

607 

719 

829 

953 

29 

103 

193 

283 

397 

499 

613 

727 

839 

967 

31 

107 

197 

293 

401 

503 

617 

733 

853 

971 

37 

109 

199 

307 

409 

509 

619 

739 

857 

977 

41 

113 

211 

311 

419 

521 

631 

743 

859 

983 

43 

127 

223 

313 

421 

523 

641 

751 

863 

991 

47 

131 

227 

317 

431 

541 

643 

757 

877 

997 

53 

137 

229 

331 

433 

547 

647 

761 

881 

FACTORING  NUMBERS. 

CASE   I. 

91.    To   resolve   any  composite    number  into   its 
prime  factors. 

Whai  is  a  prime  number  ?    In  factoring  numbers,  Case  I  is  what  ? 


76  PROPERTIES   OP   NUMBERS. 

1.    What  are  the  prime  factors  of  2772  ? 
OPERATION.          ANALYSIS.     We  divide  the  given  number  by 


2 
2 
3 
3 
7 
11 


2772  2,  the  least  prime  factor,  and  the  result  by  2; 
this  gives  an  odd  number  for  a  quotient,  divisible 
by  the  prime  factor,  3,  and  the  quotient  resulting 
from  this  division  is  also  divisible  by  3.  The 
next  quotient,  77,  we  divide  by  its  least  prime 
factor,  7,  and  we  obtain  the  quotient  11 ;  this  be- 
ing a  prime  number,  the  division  can  not  be  car- 
ried further.  The  divisors  and  last  quotient,  2, 
2,  3,  3,  7,  and  11  are  all  the  prime  factors  of  the 


given  number,  2772.     Hence  the 


RULE.  Divide  the  given  number  by  any  prime  factor  ;  di~ 
vide  the  quotient  in  the  same  manner,  and  so  continue  the 
division  until  the  quotient  is  a  prime  number.  The  several 
divisors  and  the  last  quotient  will  he  the  prime  factors  required. 

PROOF.  The  product  of  all  the  prime  factors  will  be  the 
given  number. 

EXAMPLES    FOR    PRACTICE. 

2.  What  are  the  prime  factors  of  1 1 40  ?     Ans.  2}  2,  3, 5, 19* 

3.  What  are  the  prime  factors  of  29925  ? 

4.  What  are  the  prime  factors  of  2431  ? 

5.  Find  the  prime  factors  of  12673. 

6.  Find  the  prime  factors  of  2310. 

7.  Find  the  prime  factors  of  2205. 

8.  What  are  the  prime  factors  of  13981  ? 

CASE   II. 

92.  To  resolve  a  number  into  all  the  different  sets 
of  factors  possible. 

1.   In  36  how  many  sets  of  factors,  and  what  are  they? 
Give  explanation.    Rule.    Proof.     Caac  II  is  what  ? 


CANCELLATION.  77 

OPERATION.  ANALYSIS.    Writing  the  36  at 

'2  X  18  th®  k^  °f  ^e  s*&n  =»  we  arrange 

3  NX  12  a^  the  different  sets  of  factors  into 

4.  y  o  which  it  can  be  resolved  under 
each  other,  as  shown  in  the  opera- 

36  rr<i  tion,  and  we  find  that  36  can  be 

2X2X9  resolved  into  8  sets  of  factors. 
2X3X6 
3X3X4 
2X2X3X3 

EXAMPLES    FOR   PRACTICE. 

2.  How  many  sets  of  factors  in  the  number  24?    What 
are  they  ?  Ans.   6  sets. 

3.  In  125  how  many  sets  of  factors  ?     What  are  they  ? 

Ans.   2  sets. 

4.  In  40  how  many  sets  of  factors,  and  what  are  they  ? 

Ans.   6  sets. 

5.  In  72  how  many  sets  of  factors,  and  what  are  they  ? 

Ans.    16  sets. 

CANCELLATION. 

93.  Cancellation  is  the  process  of  rejecting  equal  factors 
from  numbers  sustaining  to  each  other  the  relation  of  dividend 
and  divisor. 

It  has  been  shown  (7T)  that  the  dividend  is  equal  to  the 
product  of  the  divisor  multiplied  by  the  quotient.  Hence,  if 
the  dividend  can  be  resolved  into  two  factors,  one  of  which  is 
the  divisor,  the  other  factor  will  be  the  quotient. 

1.   Divide  63  by  7. 

OPERATION.  ANALYSIS.    We  see  in 

Divisor,         #)#  x  9      Dividend.  this  example  that  63  is 

composed  of  the  factors  7 

9        Quotient.  an(J  g?  an^  j-jjat  fae  factor 

7  is  equal  to  the  divisor. 

Therefore  we  reject  the  factor  7,  and  the  remaining  factor,  9,  is  the 
quotient. 

Give  explanation.  What  is  cancellation  ?  Upon  what  principle  is 
it  based  ?  Give  first  explanation. 


78  PROPERTIES    OF   NUMBERS. 


Whenever  the  dividend  and  divisor  are  each  composite 
numbers,  the  factors  common  to  both  may  first  be  rejected 
without  altering  the  final  result.  (875  Prin.  III.) 

2.  "What  is  the  quotient  of  24  times  56  divided  by  7  times 
48? 

OPERATION.  ANALYSIS. 

24  X  56       4  X  0  X  *  X  $  We  first  in- 

—  4,  A.US.  dicate  the  op- 

7x48  #  X  0  X  $  eration  to  be 

performed  by 

writing  the  numbers  which  constitute  the  dividend  above  a  line,  and 
those  which  constitute  the  divisor  below  it.  Instead  of  multiplying 
24  by  56,  in  the  dividend,  we  resolve  24  into  the  factors  4  and  6, 
and  56  into  the  factors  7  and  8  ;  and  48  in  the  divisor  into  the  fac- 
tors 6  and  8.  We  next  cancel  the  factors  6,  7,  and  8,  which  are 
common  to  the  dividend  and  divisor,  and  we  have  left  the  factor  4 
in  the  dividend,  which  is  the  quotient. 

NOTE.    When  all  the  factors  or  numbers  in  the  dividend  are  can- 
celed, 1  should  be  retained. 

95.    If  any  two  numbers,  one  in  the  dividend  and  one  in 
the  divisor,  contain  a  common  factor,  we  may  reject  that  factor. 

3.  In  54  times  77,  how  many  times  63  ? 

OPERATION.  ANALYSIS.     In  this  example  we  see  that  9  will 

6         11         divide  54  and  63  ;  so  we  reject  9  as  a  factor  of  54, 

^  ,        #tf         an(^  reta^n  ^e  factor  6,  and  also  as  a  factor  of  63, 

ft£  X  /iff         and  retam  the  factor  7.     Again,  7  will  divide  7  in 

0$  the   divisor,  and  77  in  the  dividend.     Dividing 

^  both  numbers   by  7,  1  will  be  retained  in  the 

divisor,   and    11   in  the  dividend.     Finally,  the 

product  of  6  X  1  1  =  66,  the  quotient. 

4.  Divide  25  X  16  X  12  by  10  X  4  X  6  X  7. 


OPERATION. 

54       $ 

ANALYSIS.  In 
this,  as  in  the  pre- 
ceding example,  we 

^0X^X0X7        7 

tors  that  are  com- 
mon to  both  divi- 
dend and  divisor, 

Give  second  explanation. 


CANCELLATION.  79 

and  we  have  remaining  the  factor  7  in  the  divisor,  and  the  factors  5  and 
4  in  the  dividend.     Completing  the  work,  we  have  ^°-  =  2£,  Ans. 

From  the  preceding  examples  and  illustrations  we  derive 
the  following 

RULE.  I.  Write  the  numbers  composing  the  dividend  above 
a  horizontal  line,  and  the  numbers  composing  the  divisor 
below  it. 

II.  Cancel  all  the  factors  common  to  both  dividend  and 
divisor. 

III.  Divide  the  product  of  the  remaining  factors  of  the  div- 
idend by  the  product  of  the  remaining  factors  of  the  divisor y 
and  the  result  will  be  the  quotient. 

NOTES.  1.  Rejecting  a  factor  from  any  number  is  dividing  the  number 
by  that  factor. 

2.  When  a  factor  is  canceled,  the  unit,  1,  is  supposed  to  take  its 
place. 

3.  One  factor  in  the  dividend  -will  cancel  only  one  equal  factor  in  the 
divisor. 

4.  If  all  the  factors  or  numbers  of  the  divisor  are  canceled,  the 
product  of  the  remaining  factors  of  the  dividend  will  be  the  quotient. 

5.  By  many  it  is  thought  more  convenient  to  write  the  factors  of 
the  dividend  on  the  right  of  a  vertical  line,  and  the  factors  of  the  divisor 
on  the  left. 

EXAMPLES    FOR -PRACTICE. 

1.   What  is  the  quotient  of  16  X  5  X  4  divided  by  20  X  8  ? 

FIRST  OPERATION.  SECOND   OPERATION. 

2  0, 

=  2, 


2,  Ans. 
2.   Divide  the  product  of  120  X  44  X  6  X  7  by  72  X  33  X  14. 

Rule,  first  step  ?  Second  ?  Third  ?  What  is  the  effect  of  rejecting 
a  factor  ?  What  is  the  quotient  when  all  the  factors  in  the  divisor  are 
canceled  ? 


80  PROPERTIES  OF  NUMBERS. 

FIRST  OPERATION. 


_ 

—  --£- 

3 


SECOND  OPERATION. 

,10 


"U 


3.  Divide  the  product  of  33  X  35  X  28  by  11  X  15  X  14. 

Am.   14. 

4.  What  is  the  quotient  of  21  X  11  X  26  divided  by  14  X 
13?  Am.   33. 

5.  Divide  the  product  of  the  numbers  48,  72,  28,  and  5,  by 
the  product  of  the  numbers  84,  15,  7,  and  6,  and  give  the 
result.  Ans.    9^-. 

'6.   Divide  140  X  39  X  13  X  7  by  30  X  7  X  26  X  21. 

Ans.   4^. 

7.  What  is  the  quotient  of  66  X  9  X  18  X  5  divided  by 
22  X  6  X  40  ?  Ans.    10J. 

8.  Divide  the  product  of  200  X  36  X  30  X  21  by  270  X 
40  X  15  X  14.  Ans.   2. 

9.  Multiply  240  by  56,  and  divide  the  product  by  60  mul- 
tiplied by  28.  Ans.   8. 

10.  The  product  of  the  numbers  18,  6,  4,  and  42  is  to  be 
divided  by  the  product  of  the  numbers  4,  9,  3,  7,  and  6 ;  what 
is  the  result  ?  Ans.   4. 

11.  How  many  tons  of  hay,  at  12  dollars  a  ton,  must  be 
given  for  30  cords  of  wood,  at  4  dollars  a  cord  ?    Ans.  10  tons. 


GREATEST   COMMON  DIVISOR.  81 

12.  How  many  firkins  of  butter,  each  containing  56  pounds, 
at  13  cents  a  pound,  must  be  given  for  4  barrels  of  sugar,  each 
containing  182  pounds,  at  6  cents  a  pound  ?     Am.    6  firkins. 

13.  A  tailor  bought  5  pieces  of  cloth,  each  piece  containing 
24  yards,  at  3  dollars  a  yard.     How  many  suits  of  clothes,  at 
18  dollars  a  suit,  must  be  made  from  the  cloth  to  pay  for  it? 

Am.   20  suits. 

14.  How  many  days'  work,  at  75  cents  a  day,  will  pay  for 
115  bushels  of  corn,  at  50  cents  a  bushel?     Ans.   76fdays. 


GREATEST  COMMON  DIVISOR. 

9O.    A  Common  Divisor  of  two  or  more  numbers  is  a 
number  that  will  exactly  divide  each  of  them. 

97.  The  Greatest  Common  Divisor  of  two  or  more  num- 
bers is  the  greatest  number  that  will  exactly  divide  each  of 
them. 

Numbers  prime  to  each  other  are  such  as  have  no  common 
divisor. 

NOTE.    A  common  divisor  is  sometimes  called  a  Common  Measure ; 
and  the  greatest  common  divisor,  the  Greatest  Common  Measure. 

CASE   I. 

98.  "When  the  numbers  are  readily  factored. 

1.  What  is  the  greatest  common  divisor  of  6  and  10  ? 

Am.   2. 

OPERATION.          ANALYSIS.    We  readily  find  by  inspection 
6  .  .  1 0         that  2  will  divide  both  the  given  numbers ; 

hence  2  is  a  common   divisor ;   and  since  the 

**  • ' _        quotients  3  and  5  have  no  common  factor,  but 

are  prime  to  each  other,  the  common  divisor, 
2,  must  be  the  greatest  common  divisor. 

2.  What  is  the  greatest  common  divisor  of  42,  63,  and  105  ? 


What  is  a  common  divisor  ?  The  greatest  common  divisor  ?  A 
common  measure  ?  The  greatest  common  measure  ?  What  is  Case  I  ? 
Give  analysis. 


82  PROPERTIES   OF  NUMBERS. 

OPERATION.  ANALYSIS.    We  observe  that  3 


14..  21  ..    35 


42  .  .  63  .  .  105  wnl  exactly  divide  each  of  the  given 

numbers,   and  that   7   will   exactly 
divide   each  of  the  resulting  quo- 
2  . .    3  . .       5  tients.     Hence,  each  of  the  given 

-~^-^ numbers  can  be  exactly  divided  by  3 

**  X  «—21,  Ans.  times  7  ;  and  these  numbers  must  be 

component  factors  of  the   greatest 

common  divisor.  Now,  if  there  were  any  other  component  factor  of 
the  greatest  common  divisor,  the  quotients,  2, 3,  5,  would  be  exactly 
divisible  by  it.  But  these  quotients  are  prime  to  each  other.  Hence 
3  and  7  are  all  the  component  factors  of  the  greatest  common  divisor 
sought. 

3.   What  is  the  greatest  common  divisor  of  28, 140,  and  280  ? 

OPERATION.  ANALYSIS,    We  first  divide  by  4  ; 

28  . .  1 40  . .  280  tnen  tne  quotients  by  7.     The  re-  • 

; suiting  quotients,  1,  5,  and  10,  are 

'  •  •    "°  •  •     *    _  prime  to  each  other.     Hence  4  and 


1  5  . .     10         7  are  all  the  component  factors  of 

the  greatest  common  divisor. 

4  X  7  =  28,  Ans. 

From  these  examples  and  analyses  we  derive  the  following 

RULE.     I.    Write  the  numbers  in  a  line,  with  a  vertical  line 
at  the  left,  and  divide  by  any  factor  common  to  all  the  numbers. 

II.  Divide  the  quotients  in  like  manner,  and  continue  the 
division  till  a  set  of  quotients  is  obtained  that  have  no  common 
factor. 

III.  Multiply  all  the  divisors  together,  and  the  product  will 
be  the  greatest  common  divisor  sought. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  12,  36,  60,  72  ? 

Ans.    12. 

2.  What  is  the  greatest  common  divisor  of  18,  24,  30,  36, 
42?  Ans.    6. 

Rule,  first  step  ?     Second  ?     Third  ? 


GREATEST   COMMON   DIVISOR.  83 

3.  What  is  the  greatest  common  divisor  of  72,  120,  240, 
384?  Ans.   24. 

4.  What  is  the  greatest  common  divisor  of  36,  126,  72, 
216?  Ans.    18. 

'  5.   What  is  the  greatest  common  divisor  of  42  and  112  ? 

Ans.    14. 

6.  What  is  the  greatest  common  divisor  of  32,  80,  and 
256?  Ans.    16. 

7.  What  is  the  greatest  common  divisor  of  210,  280,  350, 
630,  and  840  ?  Ans.   70. 

8.  What  is  the  greatest  common  divisor  of  300,  525,  225, 
and  375  ?  Ans.   75. 

9.  What  is  the  greatest  common  divisor  of  252,  630,  1134, 
and  1386?  Ans.    126. 

10.  What  is  the  greatest  common  divisor  of  96  and  544  ? 

Ans.    32. 

11.  What  is  the  greatest  common  divisor  of  468  and  1184? 

Ans.    4. 

12.  What  is  the  greatest  common  divisor  of  200,  625,  and 
150?  Ans.   25. 

CASE    II. 

99.    When  the  numbers  can  not  be  readily  factored. 

As  the  analysis  of  the  method  under  this  case  depends  upon 
three  properties  of  numbers  which  have  not  been  introduced, 
we  present  them  in  this  place. 

I.  An  exact  divisor  divides  any  number  of  times  its  dividend. 

II.  A  common  divisor  of  two  numbers  is  an  exact  divisor 
of  their  sum. 

III.  A  common  divisor  of  two  numbers  is  an  exact  divisor 
of  their  difference. 


What  is  Case  II  ?      What  is  the  first  principle  upon  which  it  is 
founded  ?     Second  ?     Third  ? 


84 


PEOPEBTIES  OP  NUMBERS. 


84 
70 

OPEE 

2 
2 

ATION. 

203 
168 

35 

14 
14 

2 
2 

28 

7, 

0 

Ans. 


1.  What  is  the  greatest  common  divisor  of  84  and  203  ? 

ANALYSIS.  We  draw  two  vertical 
lines,  and  place  the  larger  number  on 
the  right,  and  the  smaller  number  on 
the  left,  one  line  lower  down.  We 
then  divide  203,  the  larger  number,  by 
84,  the  smaller,  and  write  2,  the  quo- 
tient, between  the  verticals,  the  prod- 
uct, 168,  opposite,  under  the  greater 
number,  and  the  remainder,  35,  below. 
We  next  divide  84  by  this  remainder, 

writing  the  quotient,  2,  between  the  verticals,  the  product,  70,  on  the 
left,  and  the  new  remainder,  14,  below  the  70.  We  again  divide  the 
last  divisor,  35,  by  14,  and  obtain  2  for  a  quotient,  28  for  a  product, 
and  7  for  a  remainder,  all  of  which  we  write  in  the  same  order  as  in 
the  former  steps.  Finally,  dividing  the  last  divisor,  14,  by  the  last 
remainder,  7,  and  we  have  no  remainder.  7,  the  last  divisor,  is  the 
greatest  common  divisor  of  the  given  numbers. 

In  order  to  show  that  the  last  divisor  in  such  a  process  is 
the  greatest  common  divisor,  we  will  first  trace  the  work  in  the 
reverse  order,  as  indicated  by  the  arrow  line  below. 


OPERATION.  7  divides  the  14,  as  proved 

by  the  last  division;  it  will 
also  divide  two  times  14,  or  28, 
(I.)  Now,  as  7  divides  both 
itself  and  28,  it  will  divide  35, 
their  sum,  (II.)  It  will  also 
divide  2  times  35,  or  70,  (I ;) 
and  since  it  is  a  common 
divisor  of  70  and  14,  it  must 
divide  their  sum,  84,  which 
is  one  of  the  given  numbers, 
(II.)  It  wiU  also  divide  2 
times  84,  or  168,  (I;)  and 

since  it  is  a  common  divisor  of  168  and  35,  it  must  divide  their 
Bum,  203,  the  larger  number,  (II.)  Hence  7  is  a  common  divisor 
of  the  given  numbers. 

Again,  tracing  the  work  in  the  direct  order,  as  indicated  below,  we 


203 


168 


35 


28 


Give  analysis. 


84 
70 
14 


GREATEST   COMMON  DIYISOE.  85 

know  that  the  greatest  common  divisor,  whatever  it  be,  must  divide 
^  2  times  84,  or  168,  (I.)    Then 

203  since  it  will  divide  both  168 

and  203,  it  must  divide  their 
difference,  35,  (HI.)  It  will 
also  divide  2  times  35,  or  70, 
(I ;)  and  as  it  will  divide  both 
35  70  and  84,  it  must  divide  their 

difference,    14,  (III.)     It  will 
~  also  divide  2  times  14  or  28, 

J  (I;)    and  as  it  will  divide  both 

28  and  35,  it  must  divide  their 
difference,  7,  (III;)     hence,  it 
cannot  be  greater  than  7. 
Thus  we  have  shown, 

1st.    That  7  is  a  common  divisor  of  the  given  numbers. 
2d.   That  their  greatest  common  divisor,  whatever  it  be, 
cannot  be  greater  than  7.     Hence  it  must  be  7. 

From  this  example  and  analysis,  we  derive  the  following 

RULE.  I.  Draw  two  verticals,  and  write  the  two  numbers, 
one  on  each  side,  the  greater  number  one  line  above  the  less. 

II.  Divide  the  greater  number  by  the  less,  writing  the  quo- 
tient between  the  verticals,  the  product  under  the  dividend,  and 
the  remainder  below. 

III.  Divide  the  less  number  by  the  remainder,  the  last  divisor 
by  the  last  remainder,  and  so  on,  till  nothing  remains.     The 
last  divisor  will  be  the  greatest  common  divisor  sought. 

IV.  If  more  than  two  numbers  be  given,  first  find  the  greatest 
common  divisor  of  two  of  them,  and  then  of  this  divisor  and  one 
of  the  remaining  members,  and  so  on  to  the  last ;  the  last  common 
divisor  found  will  be  the  greatest  common  divisor  of  all  the 
given  numbers. 

NOTES.  1.  When  more  than  two  numbers  are  given,  it  is  better  to 
begin  with  the  least  two. 

2.  If  at  any  point  in  the  operation  a  prime  number  occur  as  a  re- 
mainder, it  must  be  a  common  divisor,  or  the  given  numbers  have  no 
common  divisor. 

Rule,  first  step  ?  Second  ?  Third  ?  Fourth  ?  What  relation  have 
numbers  when  their  difference  is  a  prime  number  ? 


86 


PROPERTIES   OF   NUMBERS. 


EXAMPLES   FOR   PRACTICE. 

1.     What  is  the  greatest  common  divisor  of  221  and  5512  ? 

OPERATION. 


5512 

221 

2 

442 

1092 

4 

884 

208 

1 

208 

Am.  13 

1 

13 

78 

6 

78 

~*0 

2.  Find  the  greatest  common  divisor  of  154  and  210. 

Ans.    14. 

3.  What  is  the  greatest  common  divisor  of  316  and  664? 

Ans.   4. 

4.  What  is  the  greatest  common  divisor  of  679  and  1869  ? 

Ans.   7. 

5.  What  is  the  greatest  common  divisor  of  917  and  1495  ? 

Ans.    1. 

6.  What  is  the  greatest  common  divisor  of  1313  and  4108  ? 

Ans.    13. 

7.  What  is  the  greatest  common  divisor  of  1649  and  5423  ? 

Ans.    17. 

The  following  examples  may  be  solved  by  either  of  the  fore- 
going methods. 

8.  John  has  35  pennies,  and  Charles  50 :  how  shall  they 
arrange  them  in  parcels,  so  that  each  boy  shall  have  the  same 
number  in  each  parcel  ?  Ans.   5  in  each  parcel. 

9.  A  speculator  has  3  fields,  the  first  containing  18,  the 
second  24,  and  the  third  40  acres,  which  he  wishes  to  divide 
into  the  largest  possible  lots  having  the  same  number  of  acres 
in  each ;  how  many  acres  in  each  lot  ?  Ans.   2  acres. 


MULTIPLES.  87 

10.  A  farmer  had  231  bushels  of  wheat,  and  273  bushels 
of  oats,  which  he  wished  to  put  into  the  least  number  of  bins 
containing  the  same  number  of  bushels,  without  mixing  the 
two  kinds  ;  what  number  of  bushels  must  each  bin  hold  ? 

Ans.    21. 

11.  A  village  street  is  332  rods  long;  A  owns  124  rods 
front,  B  116  rods,  and  C  92  rods;  they  agree  to  divide  their 
land  into  equal  lots  of  the  largest  size  that  will  allow  each 
one  to  form  an  exact  number  of  lots ;  what  will  be  the  width 
of  the  lots  ?  Ans.   4  rods. 

12.  The  Erie  Railroad  has  3  switches,  or  side  tracks,  of  the 
following  lengths:    3013,  2231,  and  2047  feet;  what  is  the 
length  of  the  longest  rail  that  will  exactly  lay  the  track  on 
each  switch  ?  Ans.   23  feet. 

13.  A  forwarding  merchant  has  2722  bushels  of  wheat, 
1822  bushels  of  corn,  and  1226  bushels  of  beans,  which  he 
wishes  to  forward,  in  the  fewest  bags  of  equal  size  that  will 
exactly  hold  either  kind  of  grain ;   how  many  bags  will  it 
take?  Ans.  2885. 

14.  A  has  120  dollars,  B  240  dollars,  and  C  384  dollars ; 
they  agree  to  purchase  cows,  at  the  highest  price  per  head  that 
will  allow  each  man  to  invest  all   his   money;    how  many 
cows  can  each  man  purchase?     Ans.   A  5,  B  10,  and  C  16. 

MULTIPLES. 

100.  A  Multiple   is  a  number  exactly  divisible  by  a 
given  number ;  thus,  20  is  a  multiple  of  4. 

101.  A  Common  Multiple  is  a  number  exactly  divisible 
by  two  or  more  given  numbers ;  thus,  20  is  a  common  multiple 
of  2,  4,  5,  and  10. 

1O2;  The  Least  Common  Multiple  is  the  least  number 
exactly  divisible  by  two  or  more  given  numbers ;  thus,  24  is 
the  least  common  multiple  of  3,  4,  6,  and  8. 

What  is  a  multiple  ?  A  common  multiple  ?  The  least  common 
multiple  ? 


PROPERTIES   OF  NUMBERS. 

103.  From  the  definition  (1OO)  it  is  evident  that  the 
product  of  two  or  more  numbers,  or  any  number  of  times  their 
product,  must  be  a  common  multiple  of  the  numbers.    Hence, 
A  common  multiple  of  two  or  more  numbers  may  be  found  by 
multiplying  the  given  numbers  together. 

104.  To  find  the  least  common  multiple. 

FIRST    METHOD. 

From  the  nature  of  prime  numbers  we  derive  the  follow- 
ing principles  :  — 

I.  If  a  number  exactly  contain  another,  it  will  contain  all 
the  prime  factors  of  that  number. 

II.  If  a  number  exactly  contain  two  or  more  numbers,  it 
will  also  contain  all  the  prime  factors  of  those  numbers. 

III.  The  least  number  that  will  exactly  contain  all  the 
prime  factors  of  two  or  more  numbers,  is  the  least  common 
multiple  of  those  numbers. 

1.   Find  the  least  common  multiple  of  30,  42,  66,  and  78. 

OPERATION.  ANALYSIS.     The 

30  =  2  X  3  X  5  number  cannot  be 

42  =  2  X  3  X  7  *ess  ^an  *8,  s*nce 

66  =  2X3XH  **  must  contain  78  ; 

rro  _  OWQ\X1Q  hence   it  HlUSt    COn- 

2X6X1  tain  the  factors  of 

2X3X13X11X7X5  = 


Q 
^  X  «  X  !«• 

We  here  have  all  the  prime  factors  of  78,  and  also  all  the  factors  of 
66,  except  the  factor  11.     Annexing  11  to  the  series  of  factors, 

2X  3X  13  X  11, 

and  we  have  all  the  prime  factors  of  78  and  66,  and  also  all  the 
factors  of  42  except  the  factor  7.     Annexing  7  to  the  series  of  factors, 

2  X  3  X  13  X  11  X  7, 
and  we  have  all  the  prime  factors  of  78,  66,  and  42,  and  also  all  the 

How  can  a  common  multiple  of  two  or  more  numbers  be  found  ? 
First  principle  derived  from  prime  numbers  ?  Second  ?  Third  ? 
Give  analysis. 


LEAST  COMMON  MULTIPLE.  89 

factors  of  30  except  the  factor  5.     Annexing  5  to  the  series  of  factors, 
2XSX  13  X  11  X7X5, 

and  we  have  all  the  prime  factors  of  each  of  the  given  numbers ;  and 
hence  the  product  of  the  series  of  factors  is  a  common  multiple  of 
the  given  numbers,  (II.)  And  as  no  factor  of  this  series  can  be 
omitted  without  omitting  a  factor  of  one  of  the  given  numbers,  the 
product  of  the  series  is  the  least  common  multiple  of  the  given 
numbers,  (III.) 

From  this  example  and  analysis  we  deduce  the  following 

RULE.   I.   Resolve  the  given  numbers  into  their  prime  factors. 

II.  Take  all  the  prime  factors  of  the  largest  number,  and 
such  prime  factors  of  the  other  numbers  as  are  not  found  in  the 
largest  number,  and  their  product  will  be  the  least  common 
multiple. 

NOTE.  When  a  prime  factor  is  repeated  in  any  of  the  given  numbers, 
it  must  be  used  as  many  times,  as  a  factor  of  the  multiple,  as  the 
greatest  number  of  times  it  appears  in  any  of  the  given  numbers. 

EXAMPLES    FOR   PRACTICE. 

2.  Find  the  least  common  multiple  of  7,  35,  and  98. 

Ans.   490. 

3.  Find  the  least  common  multiple  of  24,  42,  and  17. 

Ans.   2856. 

4.  What  is  the  least  common  multiple  of  4,  9,  6,  8  ? 

Ans.   72. 

5.  What  is  the  least  common  multiple  of  8,  15,  77,  385  ? 

Ans.   9240. 

6.  What  is  the  least  common  multiple  of  10,  45,  75,  90  ? 

Ans.  450. 

7.  What  is  the  least  common  multiple  of  12,  15,  18,  35  ? 

Ans.   1260. 

Rule,  first  step  ?     Second  ?     What  caution  is  given  ? 
H* 


90  PROPERTIES   OP   NUMBERS. 


SECOND    METHOD. 


1O5.     1.  What  is  the  least  common  multiple  of  4,  6,  9, 
and  12  ? 


OPERATION.  ANALYSIS.    We  first  write 

4  . .  6  . .  9  .  .  12  tiie  given  numbers  in  a  series, 

with  a  vertical  line  at  the  left. 
Since  2  is  a  factor  of  some  of 


2. .3. .9..    6 


. .  9  . .    3  the  given  numbers,  it  must  be 

a  factor  of  the  least  common 

3  multiple  sought.     Dividing  as 


2X2X3X3=:  36,  Ans.         ™any  of  the  numbers  ^as  are 

divisible  by  2,  we  write  the 

quotients  and  the  undivided  number,  9,  in  a  line  underneath.  We 
now  perceive  that  some  of  the  numbers  in  the  second  line  contain 
the  factor  2 ;  h^nce  the  least  common  multiple  must  contain  another 
2,  and  we  again  divide  by  2,  omitting  to  write  down  any  quotient 
when  it  is  1.  We  next  divide  by  3  for  a  like  reason,  and  still  again 
by  3.  By  this  process  we  have  transferred  all  the  factors  of  each 
of  the  numbers  to  the  left  of  the  vertical ;  and  their  product,  36, 
must  be  the  least  common  multiple  sought,  (1OA,  III.) 

2.   What  is  the  least  common  multiple  of  10,  12, 15,  and  75  ? 
OPERATION.  ANALYSIS.    Weread- 


2,5 

2,3 

5 


10  .  .  12  .  .  15  .  .  75  ily  see  that  2  and  5  are 

among  the  factors  of  the 
given  numbers,  and  must 
be  factors  of  the  least 


6..    3..  15 


-  common  multiple :  hence 
v  V  H  V  /)  —  SOD     Avt* 
&  A  o  A  ^  -  ouu,  s±ns.        we  diyide  ey       number 


that  is  divisible  by  either  of  these  factors  or  by  their  product  ;  thus, 
we  divide  10  by  both  2  and  5  ;  12  by  2  ;  15  by  5  ;  and  75  by  5. 
We  next  divide  the  second  line  in  like  manner  by  2  and  3  ;  and 
afterwards  the  third  line  by  5.  By  this  process  we  collect  the 
factors  of  the  given  numbers  into  groups  ;  and  the  product  of  the 
factors  at  the  left  of  the  vertical  is  the  least  common  multiple  sought. 

3.   What  is  the  least  common  multiple  of  6,  15,  35,  42, 
and  70? 


Give  explanation. 


3,7 

2,5 


LEAST  COMMON  MULTIPLE.  91 

OPERATION.  ANALYSIS.    In  this  oper- 

15  .    42  . .  70  ation  we  omit  the  6  and  35, 

because  they  are  exactly  con- 

o  . .    2  . .  10  tained  in  some  of  the  other 


3X7X2X5=r  210,  Am.        given  numbers;   thus,  6  is 

contained  in  42,  and  35  in 

70 ;  and  whatever  will  contain  42  and  70  must  contain  6  and  35. 
Hence  we  have  only  to  find  the  least  common  multiple  of  the  re- 
maining numbers,  15,  42,  and  70. 

From  these  examples  we  derive  the  following 

RULE.  I.  Write  the  numbers  in  a  line,  omitting  any  of  the 
smaller  numbers  that  are  factors  of  the  larger,  and  draw  a 
vertical  line  at  the  left. 

II.  Divide  by  any  prime  factor,  or  factors,  that  may  be  con- 
tained in  one  or  more  of  the  given  numbers,  and  write  the  quotients 
and  undivided  numbers  in  a  line  underneath,  omitting  the  1's/ 

III.  In  like  manner  divide  the  quotients  and  undivided  num- 
bers, and  continue  the  process  till  all  the  factors  of  the  given 
numbers  have  been  transferred  to  the  left  of  the  vertical.      Then 
multiply  these  factors  together,  and  their  product  will  be  the  least 
common  multiple  required. 

EXAMPLES    FOR   PRACTICE. 

4.  What  is  the  least  common  multiple  of  12,  15,  42,  and 
60?  Ans.   420. 

5.  What  is  the  least  common  multiple  of  21,  35,  and  42  ? 

Ans.   210. 

6.  What  is  the  least  common  multiple  of  25,  60,  100,  and 
125?  .  Ans.    1500. 

7.  What  is  the  least  common  multiple  of  16,  40,  96,  and 
105?  Ans.    3360. 

8.  What  is  the  least  common  multiple  of  4,  16,  20,  48,  60, 
and  72?  Ans.   720. 

9.  What  is  the  least  common  multiple  of  84,  100,  224,  and 
300?  Ans.    16800. 

Rule,  first  step  ?     Second  ?    Third  ? 


92  PROPERTIES  OP  NUMBERS. 

10.  What  is  the  least  common  multiple  of  270,  189,  297, 
243?  Ans.   187110. 

11.  What  is  the  least  common  multiple  of  1,  2,  3,  4,  5,  6,  7, 
8,9?  Ans.   2520. 

12.  What  is  the  smallest  sum  of  money  for  which  I  could 
purchase  an  exact  number  of  books,  at  5  dollars,  or  3  dollars, 
or  4  dollars,  or  6  dollars  each  ?  Ans.    60  dollars. 

13.  A  farmer  has  3  teams;  the  first  can  draw  12  barrels 
of  flour,  the  second   15   barrels,  and  the  third   18  barrels ; 
what  is  the  smallest  number  of  barrels  that  will  make  full 
loads  for  any  of  the  teams  ?  Ans.    180. 

14.  What  is  the  smallest  sum  of  money  with  which  I  can 
purchase  cows  at  $30  each,  oxen  at  $55  each,  or  horses  at 
$105  each?  Ans.   $2310. 

15.  A  can  shear  41  sheep  in  a  day,  B  63,  and  C  54 ;  what 
is  the  number  of  sheep  in  the  smallest  flock  that  would  furnish 
exact  days'  labor  for  each  of  them  shearing  alone  ? 

Ans.  15498. 

16.  A  servant  being  ordered  to  lay  out  equal  sums  in  the 
purchase  of  chickens,  ducks,  and  turkeys,  and  to  expend  as 
little  money  as  possible,  agreed  to  forfeit  5  cents  for  every  fowl 
purchased  more  than  was  necessary  to  obey  orders.     In  the 
market  he  found  chickens  at  12  cents,  ducks  at  30  cents,  and 
turkeys  at  two  prices,  75  cents  and  90  cents,  of  which  he  im- 
prudently took  the  cheaper ;  how  much  did  he  thereby  for- 
feit? Ans.  80  cents. 


CLASSIFICATION  OF  NUMBERS. 

Numbers  may  be  classified  as  follows : 

106.  I.   As  Even  and  Odd. 

107.  II.   As  Prime  and  Composite. 

What  is  the  first  classification  of  numbers  ?  What  is  an  even  num- 
ber ?  An  odd  number  ?  Second  classification  ?  A  prime  number  ? 
A  composite  number  ? 


CLASSIFICATION  OP  NUMBERS.  93 

108.  m.    As  Integral  and  Fractional. 

An  Integral  Number,  or  Integer,  expresses  whole  things. 
Thus,  281 ;  78  boys;  1000  books. 

A  Fractional  Number,  or  Fraction,  expresses  equal  parts 
of  a  thing.  Thus,  half  a  dollar;  three-fourths  of  an  hour; 
seven-eighths  of  a  mile. 

109.  IV.  As  Abstract  and  Concrete. 
HO.    V.   As  Simple  and  Compound. 

A  Simple  Number  is  either  an  abstract  number,  or  a 
concrete  number  of  but  one  denomination.  Thus,  48,  926 ; 
48  dollars,  926  miles. 

A  Compound  Number  is  a  concrete  number  whose  value  is 
expressed  in  two  or  more  different  denominations.  Thus,  32 
dollars  15  cents ;  15  days  4  hours  25  minutes ;  7  miles  82 
rods  9  feet  6  inches. 

111.    VI.   As  Like  and  Unlike. 

Like  Numbers  are  numbers  of  the  same  unit  value. 

If  simple  numbers,  they  must  be  all  abstract,  as  6,  62,  487 ; 
or  all  of  one  and  the  same  denomination,  as  5  apples,  62  ap- 
ples, 487  apples;  and,  if  compound  numbers,  they  must  be 
used  to  express  the  same  kind  of  quantity,  as  time,  distance, 
&c.  Thus,  4  weeks  3  days  16  hours;  1  week  6  days  9 
hours ;  5  miles  40  rods ;  2  miles  100  rods. 

Unlike  Numbers  are  numbers  of  different  unit  values.  Thus, 
75,  140  dollars,  and  28  miles ;  4  hours  30  minutes,  and  5 
bushels  1  peck. 

What  is  the  third  classification  ?  What  is  an  integral  number  ?  A 
fractional  number  ?  What  is  the  fourth  classification  ?  An  abstract 
number  ?  A  concrete  number  ?  What  is  the  fifth  classification  ?  A 
simple  number  ?  A  compound  number  ?  Sixth  classification  ?  What 
are  like  numbers  ?  Unlike  numbers  ? 


94  FRACTIONS. 


FRACTIONS. 

DEFINITIONS,    NOTATION,    AND    NUMERATION. 

.  If  a  unit  be  divided  into  2  equal  parts,  one  of  the 
parts  is  called  one  half. 

If  a  unit  be  divided  into  3  equal  parts,  one  of  the  parts  is 
called  one  third,  two  of  the  parts  two  thirds. 

If  a  unit  be  divided  into  4  equal  parts,  one  of  the  parts  is 
called  one  fourth,  two  of  the  parts  two  fourths,  three  of  the 
parts  three  fourths. 

If  a  unit  be  divided  into  5  equal  parts,  one  of  the  parts  is 
called  one  fifth,  two  of  the  parts  two  fifths,  three  of  the  parts 
three  fifths,  &c. 

The  parts  are  expressed  by  figures ;  thus, 


One  half  is  written       ± 
One  third  "  £ 

Two  thirds         " 
One  fourth         " 


Two  fourths       "  f 

Three  fourths    " 


One  fifth      is  written 

Two  fifths  « 

One  seventh  " 

Three  eighths  " 


Five  ninths  |- 

Eight  tenths         "         T8<y 

Hence  we  see  that  the  parts  into  which  a  unit  is  divided  take 
their  name,  and  their  value,  from  the  number  of  equal  parts 
into  which  the  unit  is  divided.  Thus,  if  we  divide  an  orange 
into  2  equal  parts,  the  parts  are  called  halves  ;  if  into  3  equal 
parts,  thirds ;  if  into  4  equal  parts,  fourths,  &c. ;  and  each 
third  is  less  in  value  than  each  half,  and  each  fourth  less  than 
each  third;  and  the  greater  the  number  of  parts,  the  less 
their  value. 

When  a  unit  is  divided  into  any  number  of  equal  parts,  one 
or  more  of  these  parts  is  a  fractional  part  of  the  whole  number, 
and  is  called  &  fraction.  Hence 

118.  A  Fraction  is  one  or  more  of  the  equal  parts  of  a 
unit. 

Define  a  fraction. 


DEFINITIONS,   NOTATION,   AND    NUMERATION.  95 

.  To  write  a  fraction,  two  integers  are  required,  one 
to  express  the  number  of  parts  into  which  the  whole  number 
is  divided,  and  the  other  to  express  the  number  of  these  parts 
taken.  Thus,  if  one  dollar  be  divided  into  4  equal  parts, 
the  parts  are  called  fourths,  and  three  of  these  parts  are 
called  three  fourths  of  a  dollar.  This  three  fourths  may  be 
written 

3  the  number  of  parts  taken. 

4  the  number  of  parts  into  which  the  dollar  is  divided. 

115.  The  Denominator  is  the  number  below  the  line. 
It  denominates  or  names  the  parts ;  and 

It  shows  how  many  parts  are  equal  to  a  unit. 

116.  The  Numerator  is  the  number  above  the  line. 
It  numerates  or  numbers  the  parts  ;  and 

It  shows  how   many  parts  are  taken  or  expressed  by 
the  fraction. 

117.  The  Terms  of  a  fraction  are  the  numerator  and  de- 
nominator, taken  together. 

118.  Fractions  indicate  division,  the  numerator  answering 
to  the  dividend,  and  the  denominator  to  the  divisor.     Hence, 

11O.  The  Value  of  a  fraction  is  the  quotient  of  the  nu- 
merator divided  by  the  denominator. 

13O.  To  analyze  a  fraction  is  to  designate  and  describe 
its  numerator  and  denominator.  Thus,  f  is  analyzed  as  fol- 
lows :  — 

4  is  the  denominator,  and  shows  that  the  unit  is  divided 
into  4  equal  parts  ;  it  is  the  divisor. 

3  is  the  numerator,  and  shows  that  3  parts  are  taken ;  it  is 
the  dividend,  or  integer  divided. 

3  and  4  are  the  terms,  considered  as  dividend  and  divisor. 

The  value  of  the  fraction  is  the  quotient  of  3  -^-  4,  or  J. 

How  many  numbers  are  required  to  write  a  fraction  ?  Why  ?  De- 
fine the  denominator.  The  numerator.  What  are  the  terms  of  a  frac- 
tion ?  The  value  ?  What  is  the  analysis  of  a  fraction  ? 


96  FRACTIONS. 

EXAMPLES  FOR  PRACTICE. 

Express  the  following  fractions  by  figures :  — - 

1.  Seven  eighths. 

2.  Three  twenty-fifths. 

3.  Nine  one  hundredths. 

4.  Sixteen  thirtieths. 

5.  Thirty-one  one  hundred  eighteenths. 

6.  Seventy-five  ninety-sixths. 

7.  Two  hundred  fifty-four  four  hundred  forty-thirds. 

8.  Eight  nine  hundred  twenty-firsts. 

9.  One  thousand  two  hundred  thirty-two  seventy-five  thou- 
sand six  hundredths. 

10.  Nine   hundred  six  two  hundred  forty-three   thousand 
eighty-seconds. 

Read  and  analyze  the  following  fractions : 

11.  A;  iV;  A;  *i;  If;  rfe;  A-5^  ill- 

12.  T9<&;  m 

TUTTIJ5 

Fractions  are  distinguished  as  Proper  and  Improper. 

A  Proper  Fraction  is  one  whose  numerator  is  less  than  its 
denominator ;  its  value  is  less  than  the  unit,  1.  Thus,  fy,  T5g, 
A»  £f  are  Pr°Per  fractions. 

An  Improper  Fraction  is  one  whose  numerator  equals  or 
exceeds  its  denominator;  its  value  is  never  less  than  the 
unit,  1.  Thus,  -f,  §,  J45-,  -^5-,  |-g,  JF8^°-  are  improper  fractions. 

122.  A  Mixed  Number  is  a  number  expressed  by  an  in- 
teger and  a  fraction;  thus,  4^-,  17£|,  9T3^  are  mixed  numbers. 

12*S.  Since  fractions  indicate  division,  all  changes  in  the 
terms  of  a  fraction  will  affect  the  value  of  that  fraction  according 
to  the  laws  of  division ;  and  we  have  only  to  modify  the  lan- 
guage of  the  General  Principles  of  Division  (8T)  by  substi- 
tuting the  words  numerator,  denominator,  find  fraction,  or  value 

"What  is  a  proper  fraction  ?  An  improper  fraction  ?  A  mixed 
number  ?  What  do  fractions  indicate  ? 


REDUCTION.  97 

of  the  fraction,  for  the  words  dividend,  divisor,  and  quotient, 
respectively,  and  we  shall  have  the  following 

GENERAL    PRINCIPLES    OF    FRACTIONS. 

124*  PRIN.  I.  Multiplying  the  numerator  multiplies  the 
fraction,  and  dividing  the  numerator  divides  the  fraction. 

PRIN.  II.  Multiplying  the  denominator  divides  the  fraction, 
and  dividing  the  denominator  multiplies  the  fraction. 

PRIN.  III.  Multiplying  or  dividing  both  terms  of  the  frac- 
tion by  the  same  number  does  not  alter  the  value  of  the  fraction. 

These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

12&.  A  change  in  the  NUMERATOR  produces  a  LIKE  change 
in  the  value  of  the  fraction  ;  but  a  change  in  the  DENOMINA- 
TOR produces  an  OPPOSITE  change  in  the  value  of  the  fraction. 

KEDUCTION. 

CASE  I. 

126.    To  reduce  fractions  to  their  lowest  terms. 

A  fraction  is  in  its  lowest  terms  when  its  numerator  and  de- 
nominator are  prime  to  each  other ;  that  is,  when  both  terms 
have  no  common  divisor. 

1.    Reduce  the  fraction  £§  to  its  lowest  terms. 

FIRST  OPERATION.  ANALYSIS.      Dividing    both 

||  —  §£  —  if  =.  ^,  Ans.        terms  of  a  fraction  by  the  same 

number  does  not  alter  the  value 

of  the  fraction  or  quotient,  (124,  III;)  hence,  we  divide  both 
terms  of  |f,  by  2,  both  terms  of  the  result,  ff,  by  2,  and  both  terms 
of  this  result  by  3.  As  the  terms  of  f  are  prime  to  each  other,  the 
lowest  terms  of  ||  are  |.  We  have,  in  effect,  canceled  all  the  fac- 
tors common  to  the  numerator  and  denominator. 

First  general  principle  ?  Second  ?  Third  ?  General  law  ?  What 
is  meant  by  reduction  of  fractions  ?  Case  I  is  what  r  What  is 
meant  by  lowest  terms?  Give  analysis. 

I 


98  FRACTIONS. 

SECOND  OPERATION.  In  this  operation  we  have  divided 

12}  48  •  _  4-    Ans.  both  terms  of  the  fraction  by  their 

greatest  common  divisor,   (971,)    and 
thus  performed  the  reduction  at  a  single  division.     Hence  the 

RULE.     Cancel  or  reject  all  factors  common  to  loth  numera- 
tor and  denominator.     Or, 

Divide  both  terms  by  their  greatest  common  divisor. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  ||J  to  its  lowest  terms.  Ans.    £. 

3.  Reduce  f  f  §  to  its  lowest  terms.  Ans.    f  . 

4.  Reduce  f  f  £  to  its  lowest  terms.  Ans.  f  £. 

5.  Reduce  §§|  to  its  lowest  terms. 

6.  Reduce  sVfe*?  to  its  lowest  terms. 

7.  Reduce  f^T  to  its  lowest  terms. 

8.  Reduce  -jtffg  to  its  lowest  terms. 

9.  Reduce  £$f  $•  to  its  lowest  terms.  Ans.  £J. 

10.  Reduce  |f  J$  to  its  lowest  terms.  Ans.  £J. 

11.  Reduce  •££/•$;  to  its  lowest  terms.  Ans.  £££. 

12.  Express  in  its  simplest  form  the  quotient  of  441  divided 
by  462.  Ans.   f£. 

13.  Express  in  its  simplest  form  the  quotient  of  189  di- 
vided by  273.  Ans.  TV 

14.  Express  in  its  simplest  form  the  quotient  of  1344  di- 
vided by  1536.  Ans.  £. 

CASE    II. 

137.    To  reduce  an  improper  fraction  to  a  whole 
or  mixed  number. 

1.    Reduce  3-  to  a  whole  or  mixed  number. 


OPERATION.  ANALYSIS.    Since 

—  324-M5  =  21T\  =  21f,  Ans.       15  fifteenths    equal 

1,324  fifteenths  are 
equal  to  as  many  times  1  as  15  is  contained  times  in  324,  which  is 
times.     Or,  since  the  numerator  is  a  dividend  and  the  denom- 


\ 


Rule.     Case  II  is  what  ?     Give  explanation. 


REDUCTION.  99 

inator  a  divisor,  (118,)  we  reduce  the  fraction  to  an  equivalent 
whole  or  mixed  number,  by  dividing  the  numerator,  324,  by  the 
denominator,  15.  Hence  the 

RULE.     Divide  the  numerator  by  the  denominator. 

NOTES.  1.  When  the  denominator  is  an  exact  divisor  of  the  numer- 
ator, the  result  will  be  a  whole  number. 

2.  In  all  answers  containing  fractions,  reduce  the  fractions  to  their 
lowest  terms. 

EXAMPLES    FOR   PRACTICE. 

2.  In  J^3-  of  a  week,  how  many  weeks?  Ans.    If-. 

3.  In  ^^-  of  a  bushel,  how  many  bushels  ?         Ans.   23f . 

4.  In  A  6 1  of  a  dollar,  how  many  dollars  ? 

5.  In  -8T7^  of  a  pound,  how  many  pounds  ?          Ans.    54£. 

6.  Reduce  -ftp  to  a  mixed  number. 

7.  Reduce  Zffi  to  a  whole  number. 

8.  Change  J^f2-  to  a  mixed  number.  Ans.    18§. 

9.  Change  ^ f  -1  to  a  mixed  number. 

10.  Change  ^%°/fl  to  a  mixed  number.        Ans.    1053f  f. 

11.  Change  ai^aa  to  a  whole  number.  Jtws.   7032. 

CASE    III. 

128.  To  reduce  a  whole  number  to  a  fraction  hav- 
ing a  given  denominator. 

1.    Reduce  46  yards  to  fourths. 

OPERATION.  ANALYSIS.     Since  in  1  yard  there  are  4  fourths, 

4g  in  46  yards  there  are  46  times  4  fourths,  which  are 

4  184  fourths  r=  ^4.     In  practice  we  multiply  46, 

the  number  of  yards,  by  4,  the  given  denominator, 
Af  ?  ^ns-         and  taking  the  product,  184,  for  the  numerator  of  a 
fraction,  and  the  given  denominator,  4,  for  the  de- 
nominator, we  have  ^^-.    Hence  we  have  the 

RULE.  Multiply  the  whole  number  by  the  given  denominator  ; 
take  the  product  for  a  numerator,  under  which  write  the  given 
denominator. 

Rule.     Case  HI  is  what  ?     Give  explanation.     Rule. 


100  FRACTIONS. 

NOTE.  A  whole  number  is  reduced  to  a  fractional  form  by  writing 
1  under  it  for  a  denominator  ;  thus,  9  - —  3.. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  25  bushels  to  eighths-of  a  bushel.       Ans.   ^ &. 

3.  Reduce  G3  gallons  to  fourths  of  a  gallon.       Ans.   5^2. 

4.  Reduce  140  pounds  to  sixteenths  of  a  pound. 

5.  In  56  dollars,  how  many  tenths  of  a  dollar  ?    Ans.   -5T6a°-. 

6.  Reduce  94  to  a  fraction  whose  denominator  is  9. 

7.  Reduce  180  to  seventy-fifths. 

8.  Change  42  to  the  form  of  a  fraction.  Ans,   ^Ji. 

9.  Change  247  to  the  form  of  a  fraction. 

10.  Change  347  to  a  fraction  whose  denominator  shall 
be  14.  Ans.  ±fja. 

CASE    IV. 

129.  To  reduce  a  mixed  number  to  an  improper 
fraction. 

1.    In  5|  dollars,  how  many  eighths  of  a  dollar  ? 
OPERATION. 

5  a  ANALYSIS.     Since  in  1  dollar  there  are  8  eighths, 

Q  in  5  dollars  there  are    5   times    8  eighths,  or   40 

eighths,  and  40  eighths  -f-  3  eighths  :=  43  eighths, 
or  4g3-.     From  this  operation  we  derive  the  following 

RULE.  Multiply  the  whole  number  by  the  denominator  of 
the  fraction  ;  to  the  product  add  the  numerator,  and  under  the 
sum  write  the  denominator. 

EXAMPLES    FOR    PRACTICE. 

2.  In  4£  dollars,  how  many  half  dollars  ?  .Ans.    f . 

3.  In  7 If  weeks,  how  many  sevenths  of  a  week  ? 

4.  In  341 J  acres,  how  many  fourths?  Ans. 

5.  Change  12T72  years  to  twelfths. 

G.    Change  SGy9^  to  an  improper  fraction.  Ans. 

7.  Reduce  21  fa  to  an  improper  fraction.         Ans. 

8.  Reduce  225i|  to  an  improper  fraction.       Ans.   - 


Case  IV  is  what  ?     Give  explanation.     Rule. 


REDUCTION.  101 


9.   In  96^%  how  many  one  hundred  twentieths  ? 

10.  In  1297B\,  how  many  eighty-fourths  ?     Ans. 

11.  What  improper  fraction  will  express  400f  $  ? 


CASE  v. 
13O.   To  reduce  a  fraction  to  a  given  denominator. 

As  fractions  may  be  reduced  to  lower  terms  by  division, 
they  may  also  be  reduced  to  higher  terms  by  multiplication  ; 
and  all  higher  terms  must  be  multiples  of  the  lowest  terms. 
(103.) 

1.  Reduce  £  to  a  fraction  whose  denominator  is  20. 

OPERATION.  ANALYSIS.    "We  first  divide  20,  the 

20  _i_  4  —  5  required  denominator,  by  4,  the  denomi- 

nator of  the  given  fraction,  to  ascertain 

_^  X  5  __  ^  ^n^        if  it  be  a  multiple  of  this  term,  4.     The 

4X5  division  shows  that  it  is  a  multiple,  and 

that  5  is  the  factor  which  must  be  em- 

ployed to  produce  this  multiple  of  4.     We  therefore  multiply  both 
terms  of  f  by  5,  (124,)  and  obtain  -^f,  the  desired  result.    Hence  the 

RULE.  Divide  the  required  denominator  by  the  denominator 
of  the  given  fraction,  and  multiply  both  terms  of  the  fraction  by 
the  quotient. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  f  to  a  fraction  whose  denominator  is  15. 

Ans.  -ft. 

3.  Reduce  f  to  a  fraction  whose  denominator  is  35. 

4.  Reduce  Jf  to  a  fraction  whose  denominator  is  51. 

Ans.   ff. 

5.  Reduce  f  §  to  a  fraction  whose  denominator  is  150. 

6.  Reduce  £§|  to  a  fraction  whose  denominator  is  3488. 

Ans. 

7.  Reduce          to  a  fraction  whose  denominator  is  1000. 


Case  V  is  what  ?     How  are  fractions  reduced  to  higher  terms  ? 
What  are  all  higher  terms  ?     Give  analysis.     Rule. 

T* 


102  FRACTIONS. 

CASE   VI. 

131.    To  reduce  two  or  more  fractions  to  a  com- 
mon denominator. 

A  Common  Denominator  is  a  denominator  common  to  two 
or  more  fractions. 

1.  Reduce  f  and  f  to  a  common  denominator. 

OPERATION.  ANALYSIS.    We  multiply  the  terms  of  the 

3><5  first  fraction  by  the  denominator  of  the  second, 

—           =  £$  and  the  terms  of  the  second  fraction  by  the 

denominator  of  the  first,  (124.)  This  must  re- 

2  <x  ^  duce  each  fraction  to  the  same  denominator, 

—   8  for  each  new  denominator  will  be  the  product 

5X4  of  the  given  denominators.     Hence  the 

RULE.    Multiply  the  terms  of  each  fraction  by  the  denomina- 
tors of  all  the  other  fractions. 

NOTE.   Mixed  numbers  must  first  be  reduced  to  improper  fractions. 
EXAMPLES    FOR    PRACTICE. 

2.  Reduce  §,  £,  and  J  to  a  common  denominator. 

Ans.   «,Jf,£. 

3.  Reduce  f  and  |  to  a  common  denominator. 

>  Ans.    §5,  f|. 

4.  Reduce  f,  -f^  and  £  to  a  common  denominator. 

Ans.   HI,  SJg,  m- 

5.  Reduce  f ,  §,  f ,  and  ^  to  a  common  denominator. 

Ans.   Mi,  f4J,  Ml,  MI- 
G.    Reduce  ^5,  £,  and  f  to  a  common  denominator. 

Ans.    mifo£h- 

7.  Reduce  f ,  2£,  J,  and  £  to  a  common  denominator. 

A*"-  -HI,  ?«»  iif'  Aft- 

8.  Reduce  1 J,  ^0,  and  4  to  a  common  denominator. 

Ans. 


C.ise  TI  ic  wlmt  ?     "What  is  a  common  denominator  ?     Give  analysis. 


EEDUCTION.  103 

CASE  VII. 

To  reduce  fractions  to  the  least  common  de- 
nominator. 

The  Least  Common  Denominator  of  two  or  more  fractions 
is  the  least  denominator  to  which  they  can  all  be  reduced,  and 
it  must  be  the  least  common  multiple  of  the  lowest  denom- 
inators. 

1.   Eeduce  £,  f,  and  T52  to  the  least  common  denominator. 
OPERATION.  ANALYSIS.    We  first  find 


g      -^2  the  least  common  multiple 

"  .  '      g 

"    ' 


2, 3      „  ..  ^  

Q'  a  4.9  °^  t^ie  S^ven  denominators, 

III "    '  which  is  24.     This  must  be 


2X3X2X2  —  24  the  least  common  denom- 

i   —   4   ^  inator   to  which  the  frac- 

3  —  _9    I  Ans.         tions  can  be  reduced.  (III.) 

f_  __  ^J  j  We  then  multiply  the  terms 

of  each  fraction  by  such  a 

number  as  will  reduce  the  fraction  to  the  denominator,  24.  Re- 
ducing each  fraction  to  this  denominator,  by  Case  V,  we  have  the 
answer. 

»$ince  the  common  denominator  is  already  determined,  it  is 
only  necessary  to  multiply  the  numerators  by  the  multipliers. 
Hence  the  following 

KULE.  I.  Find  the  least  common  multiple  of  the  given  de- 
nominators, for  the  least  common  denominator. 

II.  Divide  this  common  denominator  by  each  of  the  given 
denominators,  and  multiply  each  numerator  by  the  correspond- 
ing quotient.  The  products  will  be  the  new  numerators. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  2\,  ^,  f  g,  and  T\  to  their  least  common  de- 
nominator. Ans.   TV20,  T¥^  T*£>  rfir- 

3.  Reduce  £,  ^-,  T3F,  ?2T  to  their  least  common  denominator. 

iSI,  M 


What  is  Case  VLT  ?    What  must  be  the  least  common  denominator  ? 
Give,  analysis.    Rule,  first  step.     Second. 


104  FRACTIONS. 

4.  Reduce  f,  2^,  f,  and  6  to  their  least  common  denomi- 
nator. Ans.    3&Wft>Mf,W- 

5.  Reduce  5j,  2£,  and  If  to  their  least  common  denomina- 
tor. Ans.   -4/->  J/-,  -V-- 

6.  Reduce  T9T,  f  ,  f  ,  and  £  to  their  least  common  denomi- 
nator. Ans.    I?!,  §?4,  Ml,  4ft- 

7.  Reduce  j-  ,  £,  f  ,  2|,  and  ^-  to  their  least  common  de- 
nominator. Ans.   -Jf  f,  T¥S>  T¥S»  ret,  i%- 

8.  Change  f,  T\,  3f,  9,  and  J-  to  equivalent  fractions  hav- 
ing the  least  common  denominator. 

9.  Change  §•£,  If,  |,  J|-,and  6  to  equivalent  fractions  hav- 
ing the  least  common  denominator. 

10.  Change  2r7a,  |J,  4,  If,  |£,  and  |  to  equivalent  frac- 
tions having  the  least  common  denominator. 

11.  Reduce  ^,  §,  £,  and  j7^  to  a  common  denominator. 

12.  Reduce  £,  ^,  2£,  and  ^  to  a  common  denominator. 

13.  Reduce  -£f,  T7^,  f  ,  and  3-1  to  equivalent  fractions  hav- 
ing a  common  denominator.  Ans.    |§,  §£,  §§-,  |§. 

14.  Change  /T,  f  ,  and  |^  to  equivalent  fractions  having  a 
common  denominator.  Ans.    &%%,  ^o^?  ^^< 

15.  Change  ^T,  7J-,  §§,  and  5  to  equivalent  fractions  haV- 
ing  a  common  denominator.  Ans.    f  |,  ^5-,  f§,  -3F3&0-. 

16.  Change  ^  6^,  T9^,  7,  f,  and  1^-  to  equivalent  fractions 
having  a  common  denominator. 

ADDITION. 

133.    1.   What  is  the  sum  of  £,  f  ,  f  ,  and  £  ? 

OPERATION.  ANALYSIS.     Since  the 

—  -yi—  2,  ^rzs.          ^ven   ^actions  have  a 


common  denominator,  8, 
their  sum  may  be  found  by  adding  their  numerators,  1,  3,  5,  and 
7,  and  placing  the  sum,  16,  over  the  common  denominator.  We 
thus  obtain  ^  =  2,  the  required  sum. 

2.  Add  ^,  A,  ^  A,  and  TV  Jws.    2J. 

3.  Add 


Give  first  explanation. 


ADDITION. 


4.  What  is  the  sum  of  T 

A 

5.  What  is  the  sum  o 

6.  What  is  the  sum  o 


,  A  H> 
Jfo 
£  JJ, 


105 

IVEBSIT' 

our        «  K 
ltvn^ 

and  f  Jf  ? 


134.    1.   What  is  the  sum  off  and  f? 

OPERATION.  ANALYSIS.  In 

whole    numbers 
we  can  add  like 

numbers  only,  or  those  having  the  same  unit  value  ;  so  in  fractions 
we  can  add  the  numerators  when  they  have  a  common  denominator, 
but  not  otherwise.  As  |  and  -f  have  not  a  common  denominator, 
we  first  reduce  them  to  a  common  denominator,  and  then  add  the  nu- 
merators, 27  -|-  10  ==  37,  the  same  as  whole  numbers,  and  place  the 
sum  over  the  common  denominator.  Hence  the  following 

RULE.  I.  When  necessary,  reduce  the  fractions  to  a  com- 
mon or  to  their  least  common  denominator. 

II.  Add  the  numerators,  and  place  the  sum  over  the  common 
denominator. 

NOTE.  If  the  amount  be  an  improper  fraction,  reduce  it  to  a  whole 
or  a  mixed  number. 


EXAMPLES    FOR   PRACTICE. 

2.  Add  f  to  f.  Ans.   f  |. 

3.  Addf  to|i.  Ans.    1 

4.  Add  f  ,  |,  f  and  TV.  Ans. 

5.  Add  if,  Sfc  and  ^-.  Ans. 

6.  Add  ^,  A,  /y,  and  A.  Ans.  f. 

7.  Add  fj,  if  *,  ||,  3,  and  f  .  ^w.   3 

8.  Add  f  ,  J,  f,  t,  $,  f  J,  f  ,  and  &.  -4n». 

9.  Add  7J-,  of,  and  10}. 

OPERATION. 
f  +  f+    f—    1|£ 
7  _L  5  _l_  10  ^^  22 


Ans.    23  }£ 


ANALYSIS.  The  sum  of  the  frac- 
tions^,f,andf  is  1^;  the  sum  of 
^e  integers,  7,  5,  and  10,  is  22  ; 
and  the  sum  of  both  fractions 
and  integers  is  23}J.  Hence, 


Give  second  explanation  ?    Rule,  first  step.     Second. 


106 


FRACTIONS. 


To  add  mixed  numbers,  add  the  fractions  and  integers  sep- 
arately, and  then  add  their  sums. 

NOTE.   If  the  mixed  numbers  are  small,  they  may  be  reduced  to  im- 
proper fractions,  and  then  added  after  the  usual  method. 

10.  What  is  the  sum  of  14|,  3r%  If,  and  £§?     Ans.    21JJ. 

11.  What  is  the  sum  of  },  l/^-,  lOf  ,  and  5  ?       Ans.   18/¥. 

12.  What  is  the  sum  of  17f  ,  18&,  and  26^  ? 

13.  What  is  the  sum  of  &,  ^  1  £,  3,  and  ±%  ? 

14.  What  is  the  sum  of  125f-,  327  &,  and  25£?  Ans.   478/T. 

15.  What  is  the  sum  of  JJ&,  |J,  1^,  «,  and  Jg$  ? 


16.  What  is  the  sum  of  3T9^,  2£f,  40f,  and  10TV  ? 

17.  Bought  3  pieces  of  cloth  containing  125},  96},  and 
48|  yards  ;  how  many  yards  in  the  3  pieces  ? 

18.  If  it  take  5£  yards  of  cloth  for  a  coat,  3£  yards  for  a 
pair  of  pantaloons,  and  }  of  a  yard  for  a  vest,  how  many  yards 
will  it  take  for  all  ?  Ans.    9-^. 

19.  A  farmer  divides  his  farm  into  5  fields;  the  first  con- 
tains 26^3-  acres,  the  second  40j-f  acres,  the  third  51  f-  acres, 
the  fourth  59f  acres,  and  the  fifth  62  §  acres  ;  how  many  acres 
in  the  farm?  Ans.   241  if. 

20.  A  speculator  bought  175f  bushels  of  wheat  for  205  £ 
dollars,  325f-  bushels  of  barley  for  296f  dollars,  270|f  bush- 
els of  corn  for  200^  dollars,  and  437^  bushels  of  oats  for 
156J  |  dollars  ;  how  many  bushels  of  grain  did  he  buy,  and  how 
much  did  he  pay  for  the  whole  ?  ^m    f  1209^9F  bushels. 

*  1  850£f  dollars. 

SUBTRACTION. 

135.    1.    From'  A  take  A- 

OPERATION.  ANALYSIS.    Since  the  given 

•J^  —  -fa  —  -fa  —  ^  Ans.         fractions  have  a  common  denom- 

inator, 10,  we  find  the  difference 
by  subtracting  3,  the  less  numerator,  from  7,  the  greater,  and  write 

How  aro  mixorl  mimlvr-  \\:\t\-  -.1  ?       (Jive  note. 


SUBTRACTION.  107 

the%  remainder,  4,  over  the  common  denominator,  10.    "We  thus 

obtain  T^-  =  f  ,  the  required  difference. 

2.  From  f  take  f  .  Ans.   £. 

3.  From  f  J  take  T£.  Ans.   £. 

4.  From  f  £  take  /T. 

5.  From  f  §  take  f  £. 

6.  From  -/^  take  TyF. 

7.  From  iff  take  JJf.  Ans.  ft. 

136.    1.   From  |  take  f  . 

OPERATION.  ANALYSIS. 

f  -  1  =  f  §  -  f  ft  =  3J?3V3J>  =  ft  =  A,  ^-  As  ^  whole 

numbers,  we 

can  subtract  like  numbers  only,  or  those  having  the  same  unit  value, 
so,  we  can  subtract  fractions  only  when  they  have  a  common  de- 
nominator. As  f  and  f  have  not  a  common  denominator,  we  first 
reduce  them  to  a  common  denominator,  and  then  subtract  the 
less  numerator,  30,  from  the  greater,  32,  and  write  the  difference,  2, 
over  the  common  denominator,  36.  We  thus  obtain  ^  =  T^,  the 
required  difference.  Hence  the  following 

RULE.  I.  When  necessary,  reduce  the  fractions  to  a 
common  denominator. 

II.  Subtract  the  numerator  of  the  subtrahend  from  the 
numerator  of  the  minuend,  and  place  the  difference  over  the 
common  denominator. 

EXAMPLES   FOR   PRACTICE. 


2. 

From  £  take  f  . 

Ans.  T5?. 

3. 

From  £f  take  f  . 

-4ws.   ¥9<y. 

4. 

Subtract  T4T  from  |. 

-4?15.    y1^. 

5. 

Subtract  -^  from  T8^. 

-4^5.     4^' 

6. 

Subtract  ^£  from  |f  §§. 

Ans.  ^5¥. 

7. 

Subtract  ^5%  from  f  §. 

^^s-   yWff- 

8. 

What  is  the  difference  between  9 

^  and  2f  ? 

Give  explanations.     Riile,  first  step.     Second. 


108  FRACTIONS. 

OPERATION.  ANALYSIS.    We  first  reduce    the   frac- 

9£  =  9^y  tional  parts,  J  and  f ,  to  a  common  denom- 

g3  —  2-C  inator,  12.      Since  we  cannot  take  ^  from 

^,  we  add  1  r=  ^|  to  T%,  which  makes  ^-f , 

6/2-  ^715.          and  T%  from  if  leaves  T^.     We  now  add  1 

to  the  2  in  the  subtrahend,  (5O,)    and  say» 

3  from  9  leaves  6.     We  thus  obtain  6T7^,  the  difference  required. 

Hence,  to  subtract  mixed  numbers,  we  may  reduce  the 
fractional  parts  to  a  common  denominator,  and  then  subtract 
the  fractional  and  integral  parts  separately.  Or, 

We  may  reduce  the  mixed  numbers  to  improper  fractions, 
and  subtract  the  less  from  the  greater  by  the  usual  method. 

9.    From  8£  take  3}.  Ans.    4|f. 

10.  From  2o£  take  9TV  Ans.    16T\. 

11.  From  4f  take  £f. 

12.  Subtract  1|  from  6. 

13.  Subtract  120^  from  450£.  Ans.   330 1|. 

14.  Subtract  T4^\  from  3TV  Ans.    334T\. 

15.  Find  the  difference  between  49  and  75£. 

16.  Find  the  difference  between  227|  and  196f. 

17.  From  a  cask  of  wine  containing  31£  gallons,  17f  gal- 
lons were  drawn ;  how  many  gallons  remained  ?  Ans.    13$. 

18.  A  farmer,  having  450^  acres  of  land,  sold  304|  acres ; 
how  many  acres  had  he  left  ?  Ans.    145i§. 

19.  If  flour  be  bought  for  6£  dollars  per  barrel,  and  sold 
for  7§  dollars,  what  will  be  the  gain  per  barrel  ? 

20.  From  the  sum  of  f  and  3£  take  the  difference  of  4£ 
and  5£.  Ans.   3|f. 

21.  A  man,  having  25}  dollars,  paid  6£  dollars  for  coal,  2£ 
dollars  for  dry  goods,  and  j  of  a  dollar  for  a  pound  of  tea ; 
how  much  had  he  left  ?  Ans. 

22.  What  number  added  to  2f  will  make  7£  ?  Ans. 

23.  What  fraction  added  to  |£  will  make  £§  ?    Ans. 


In  how  many  ways  may  mixed  numbers  be  subtracted  ?     What  arc 
they? 


MULTIPLICATION.  109 

24.  A  gentleman,  having  2000  dollars  to  divide  among  his 
three  sons,  gave  to  the  first  91 2  £  dollars,  to  the  second  545£ 
dollars,   and    to  the    third  the  remainder  ;   how    much    did 
the  third  receive  ?  Am.    $542TV 

25.  Bought  a  quantity  of  coal  for  136-j-9^  dollars,  and  of 
lumber  for  350f  dollars.     I  sold  the  coal  for  184|  dollars,  and 
the  lumber  for  41 6  J  dollars.     How  much  was  my  whole  gain? 

Ans. 

MULTIPLICATION. 

CASE   I. 

137.    To  multiply  a  fraction  by  an  integer. 

1.  If  1  yard  of  cloth  cost  f  of  a  dollar,  how  much  will  5 
yards  cost  ? 

OPERATION.  ANALYSIS.     Since  1  yard  cost 

3  NX  5  — -  15  —  33    ^ns  3  fourths  of  a   dollar,  5  yards 

will  cost  5  times  3  fourths  of  a 

dollar,  or  15  fourths,  equal  to  3f  dollars.     A  fraction  is  multiplied 
by  multiplying  its  numerator,  (124.) 

2.  If  1  gallon  of  molasses  cost  /^  of  a  dollar,  how  much 
will  5  gallons  cost  ? 

OPERATION.  ANALYSIS.       Since    5,   the 

T/Q-  X  5  =  -  =  1^    Ans.  multiplier,  is  a  factor  of  20,  the 

denominator,  of  the  multipli- 
cand, we  perform  the  multiplication  by  dividing  the  denominator, 
20,  by  the  multiplier,  5,  and  we  have  •£-,  equal  to  If  dollars.  A 
fraction  is  multiplied  by  dividing  its  denominator,  (124.)  Hence, 

Multiplying  a  fraction  consists  in  multiplying  its  numerator, 
or  dividing  its  denominator. 

NOTE.     Always  divide  the  denominator  when  it  is  exactly  divisible 
by  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

3.  Multiply  ^  by  5.  Ans.    \f  =  2f 

4.  Multiply  T3T  by  7.  Ans.    1|£. 

Case  I  is  what  ?     Give  explanations.  Deduction. 
J 


110  FRACTIONS. 

5.  Multiply  T9¥  by  12.  Ans.   7f  . 

6.  Multiply  /T  by  63.  Ans.    15. 

7.  Multiply  5£  by  9. 

OPERATION.  ANALYSIS.     In  multiply- 

5,1  ing  a  mixed  number,  we  first 

9"  Qr  multiply  the  fractional  part, 

5  1  __  11  and  then  the  integer,   and 

4£       JLJL  s/  9  —  9  9  _  ^Q  j  add  the  two  products  ;  or  we 

45  reduce  the  mixed  number  to 

4q  an    improper    fraction,    and 

*  then  multiply  it. 

8.  Multiply  7f  by  12.  Ans.   9  If 

9.  Multiply  TyT  by  8.  ^n».   5&3T. 

10.  Multiply  Tfe  by  51.  Ans.         2. 

11.  Multiply  15|  by  16.  Ans.     250. 

12.  Multiply  Jjy.  by  22.  ^ws.      16|. 

13.  If  a  man  earn  8T9^  dollars  a  week,  how  many  dollars 
will  he  earn  in  12  weeks  ? 

14.  What  will  9  yards  of  silk  cost  at  ^  of  a  dollar  per 
yard  ? 

15.  What  will  27  bushels  of  barley  cost  at  £  of  a  dollar 
per  bushel  ?  Ans.   23|  dollars. 

CASE   II. 

138.    To  multiply  an  integer  by  a  fraction. 

1.   At  75  dollars  an  acre,  how  much  will  f  of  an  acre  of 
land  cost  ? 

FIRST   OPERATION.  ANALYSIS.      3  fifths  of  an 

5  )  75     price  of  an  acre.  acre  will  cost  three  times  as 

much  as  1  fifth  of  an  acre. 

15     cost  of  £  of  an  acre.  Dividing  75  dollars  by  5,  we 

have  15  dollars,  the  cost  of 


Ans    45       «     "  £  "  «    «  i  of  an  acre>  which  we  mul- 

tiply  by  3,  and   obtain  45 

dollars,  the  cost  of  f  of  an  acre. 

Explain  the  process  of  multiplying  mixed  numbers.     What  is  Case 
II  ?     Give  first  explanation. 


MULTIPLICATION. 


Ill 


Or,  multiplying  the  price 
of  1  acre  by  3,  we  have  the 
cost  of  3  acres ;  and  as  % 
of  3  acres  is  the  same  as 
-|  of  1  acre,  we  divide  the 
cost  of  3  acres  by  5,  and 
we  have  the  cost  of  ^  of  an 
acre,  the  same  as  in  the  first 
operation.  Hence, 

Multiplying  by  a  fraction  consists  in  multiplying  by  the  nu- 
merator and  dividing  by  the  denominator  of  the  multiplier. 


SECOND   OPERATION. 
75    price  of  1  acre. 

3 

5  )  225    cost  of  3  acres. 
AnS.    45      "      "   f   of  an  acre. 


15 


NOTE.  By  using  the  vertical  line  and 
cancellation,  we  shall  shorten,  and  com- 
bine both  operations  in  one. 


45,  Ans. 


EXAMPLES    FOR    PRACTICE. 


2.  Multiply  3  by  f . 

3.  Multiply  100  by  T9¥. 

4.  Multiply  105  by  & 
5  Multiply  19  by  £f 
6.  Multiply  24  by  6|. 


OPERATION. 


24 


=     o 


144 
159,   Ans. 


Or,* 


7.  Multiply  42  by  9£. 

8.  Multiply  80  by  14-^. 

9.  Multiply  156  by  f£. 

10.   At  8  dollars  a  bushel,  what  will 
seed  cost  ? 


Ans.  H. 

Ans.  64f 

Ans.  85. 

Ans.  5|f. 


ANALYSIS.  We 
multiply  by  the  in- 
teger and  fraction 
separately,and  add 
the  products ;  or, 
reduce  the  mixed 
number  to  an  im- 
proper  fraction, 
and  then  multiply  by  it. 

Ans.    409£. 
Ans.    1165. 
Ans.      108. 
^  of  a  bushel  of  clover 


Give  second  explanation.     Note.     Deduction. 


112  FRACTIONS. 

11.  If  a  man  travel  36  miles  a  day,  how  many  miles  will 
he  travel  in  lOf  days  ?  Am.   384  miles. 

12.  If  a  village  lot  be  worth  450  dollars,  what  is  T7?  of  it 
worth?  Ans.    262^-  dollars. 

13.  At  16  dollars  a  ton,  what  is  the  cost  of  2J  tons  of  hay  ? 

CASE   III. 

139.   To  multiply  a  fraction  by  a  fraction. 
1.    At  f  of  a  dollar  per  bushel,  how  much  will  f  of  a  bushel 
of  corn  cost  ? 

OPERATION.  ANALYSIS. 

1st  step,          |   _i_  4  —  A,  cost  of  £  of  a  bushel.  Since  Ibush- 

2d  step,  A  X  3  =  A,      "  "   f  "   "      " 


Or,        $ 


o 


$  cost  f  times  f  of  a  dollar,  or  3  times 

£  of  f  of  a  dollar.     Dividing  f  of  a 

3  _  dollar  by  4,  we  have  T2^,  the  cost  of 

-I  _  ,      .|w-        \  of  a  bushel.     A  fraction  is  di- 

vided by  multiplying  its  denomina- 


tor, (124.)  Multiplying  the  cost  of  \  of  a  bushel  by  3,  we  have  -£% 
of  a  dollar,  the  cost  of  f  of  a  bushel.  It  will  readily  be  seen  that  we 
have  multiplied  together  the  two  numerators,  2  and  3,  for  a  new 
numerator,  and  the  two  denominators,  3  and  4,  for  a  new  denom- 
inator, as  shown  in  the  whole  work  of  the  operation.  Hence,  for 
multiplication  of  fractions,  we  have  this  general 

RULE.  I.  Reduce  all  integers  and  mixed  numbers  to 
improper  fractions. 

II.  Multiply  together  the  numerators  for  a  new  numerator, 
and  the  denominators  for  a  new  denominator. 

NOTE.    Cancel  all  factors  common  to  numerators  and  denominators. 

EXAMPLES  FOR   PRACTICE. 

2.  Multiply  f  by  J.  Ans.  ^. 

3  Multiply  I  by  \ .  Ans.  TV 

4.  Multiply  ££  by  f  f .  Ans.  f^. 

5.  Multiply  4£  by  f .  Ans.  3f . 

"What  is  Case  III  ?  Give  explanation.  Rule,  first  step  ?  Second  ? 
What  shall  be  done  with  common  factors  ? 


MULTIPLICATION.  113 

6.  What  is  the  product  of  y9^,  f  ,  f  ,  and  £  ?      Ans.      •£%. 

7.  What  is  the  product  of  If,  f  ,  2,  and  5£  ?     ^ws.  llff 

8.  What  is  the  product  of  £  of  -&,  £  of  f  of  £,  and  f  of 


. 

O 


OPERATION.  Or, 

6     #     0     *     7~~30' 


NOTE.   Fractions  with  the  word  o/  between  them 
are  sometimes  called  compound  fractions.     The  word 
of  is  simply  an  equivalent  for  the  sign  of  multiplica- 
tion, and  signifies  that  the  numbers  between  which  \      _~- 
it  is  placed  are  to  be  multiplied  together.                            «^U  |  I  —  -g-0. 

9.  Multiply  T\  of  2£  by  £  of  7£.  Ans.    Hf. 

10.  Multiply  f  of  16  by  ^  of  26f. 

11.  What  is  the  product  of  3,  ^  of  £,  and  f  oi 

12.  What  is  the  value  of  2£  times  f  of  f  of  1£  ?       Jws.    2. 

13.  What  is  the  value  of  £  of  £  of  If  times  f  of  8  ? 

14.  What  is  the  product  of  12^  multiplied  by  5£  times  6f  ? 


1-5.   At  f  of  a  dollar  per  yard,  what  will  f  of  a  yard  of 
cloth  cost  ?  Ans.   ^  of  a  dollar. 

16.  If  a  man  own  ^  of  a  vessel,  and  sell  §  of  his  share, 
what  part  of  the  whole  vessel  will  he  sell  ? 

17.  When  oats  are  worth  £  of  a  dollar  per  bushel,  what  is 
f  of  a  bushel  worth  ? 

18.  What  will  7f  pounds  of  tea  cost,  at  f  of  a  dollar  per 
pound  ?  Ans.    4^-§  dollars. 

19.  What  is  the  product  of  9f  by  4§  ? 

9f 

42  23         2 


39    product  by  4.         Or,  9?  X4|  =  -  X—  =  46. 

6f        «        «    f. 


46          "        «  4§. 


What  does  "  o/"'  signify  when  placed  between  two  fractions  ?     What 
is  a  compound  fraction  ? 
* 


114  FRACTIONS. 

To  multiply  mixed  numbers  together  we  may  either  mul- 
tiply by  the  integer  and  fractional  part  separately,  and  then 
add  their  products ;  or,  we  may  reduce  both  numbers  to 
improper  fractions,  and  then  multiply  as  in  the  foregoing  rule. 

20.  Multiply  12£  by  8£.  Ans.    108|. 

21.  What  cost  6 1  cords  of  wood,  at  2f  dollars  a  cord  ? 

22.  What  cost  }  of  2£  tons  of  hay,  at  11-&  dollars  a  ton  ? 

Ans.    $21T3F. 

23.  What  will  8f  cords  of  wood  cost,  at  2f  dollars  per 
cord?  Ans.   22^  dollars. 

24.  What  must  be  paid  for  f  of  6£  tons  of  coal,  at  £  of  7£ 
dollars  per  ton  ? 

25.  A  man  owning  $  of  a  farm,  sold  £  of  his  share ;  what 
part  of  the  whole  farm  had  he  left  ?  Ans.   £f. 

26.  Bought  a  horse  for  125J  dollars,  and  sold  him  for  £  of 
what  he  cost ;  how  much  was  the  loss  ?  Ans.    $25^. 

27.  A  owned  f  of  123|  acres  of  land,  and  sold  f  of  his 
share ;  how  many  acres  did  he  sell  ?  Ans.   49 T8^. 

28.  If  a  family  consume  1£  barrels  of  flour  a  month,  how 
many  barrels  will  five  such  families  consume  in  4T9TI  months  ? 

DIVISION. 

CASE   I. 

14O.   To  divide  a  fraction  by  an  integer. 

1.  If  my  horse  eat  T9<y  of  a  ton  of  hay  in  3  months,  what 
part  of  a  ton  will  last  him  1  month  ? 

OPERATION.  ANALYSIS.    If  he  eat  T97  of  a  ton  in 

^  -L-  3  =  T^j,  Ans.         3  months,  in  1  month  he  will  eat  ^  of 

T97  of  a  ton,  or  T97  divided  by  3.     Since 

a  fraction  is  divided  by  dividing  its  numerator,  (124,)  we  divide 
the  numerator  of  the  fraction,  -t%,  by  3,  and  we  have  ^,  the  answer. 

2.  If  3  yards  of  ribbon  cost  f  of  a  dollar,  what  will  1  yard 
cost? 

Case  I  is  what  ?     Give  first  explanation. 


DIVISION.  115 

OPERATION.  ANALYSIS.     Here  we  cannot  exactly 

5  _._  3  __    5     ^ng  divide  the  numerator  by  3 ;  but,  since  a 

fraction  is  divided  by  multiplying  the 

denominator,  (124,)  we  multiply  the  denominator  of  the  fraction, 
£,  by  3,  and  we  have  T\,  the  required  result.     Hence, 

Dividing  a  fraction  consists  in  dividing  its  numerator,  or 
multiplying  its  denominator. 

NOTE.    We  divide  the  numerator  when  it  is  exactly  divisible  by  the 
divisor ;  otherwise  we  multiply  the  denominator. 

EXAMPLES    FOR   PRACTICE. 

3.  Divide    f     by    2.  Ans.     f 

4.  Divide  ?9T    by    3.  Ans.     f. 

5.  Divide   ||    by    5.  Ans.  ff. 

6.  Divide  TW  by  25. 

7.  Divide  tf    by  14.  Ans.  £fo. 

8.  Divide  |i  by  21.  Ans.  ¥VV 

9.  If  6  pounds  of  sugar  cost  f  of  a  dollar,  how  much  will 
1  pound  cost  ? 

10.  At  7  dollars  a  barrel,  what  part  of  a  barrel  of  flour  can 
be  bought  for  f  of  a  dollar  ?  Ans.    £. 

11.  If  a  yard  of  cloth  cost  5  dollars,  what  part  of  a  yard 
can  be  bought  for  f  of  a  dollar  ?  Ans.   ^ . 

12.  If  9  bushels  of  barley  cost  1\  dollars,  how  much  will  1 
bushel  cost  ? 

OPERATION.  NoTE.    We  reduce  the  mixed  number 

71  — -  _3_6_  to  an  improper  fraction,  and  divide  as 

4-  +  $=%,  Ans.          before' 

13.  If  12  barrels  of  flour  cost  76f  dollars,  how  much  will 
1  barrel  cost  ? 

OPERATION.  ANALYSIS.     Here  we  first  divide  as  in 

12  )  76^  simple  numbers,  and  we  have  a  remainder 

of  4^.     We  reduce  this  remainder  to  an 

6|-,  Ans.      improper  fraction,  %£.,  which  we  divide  (as 

in  Ex.  1,)  and  annex  the  result,  f ,  to  the  partial  quotient,  6,  and 

WG  have  6^,  the  required  result. 

Give  second  explanation.    Deduction. 


116  FRACTIONS. 

14.  How  many  times  will  16  j  gallons  of  cider  fill  a  vessel 
that  holds  3  gallons  ?  Ans.   5&. 

15.  If  9  men  consume  f  of  9|  pounds  of  meat  in  a  day, 
how  much  does  each  man  consume  ?        Ans.   f  of  a  pound. 

16.  A  man  paid  $99f£  for  4  cows;  how  much  was  that 
apiece?  Ans.   $24ff. 

CASE   II. 

141.   To  divide  an  integer  by  a  fraction. 

1.   At  f  of  a  dollar  a  yard,  how  many  yards  of  cloth  can  be 
bought  for  12  dollars? 

FIRST  OPERATION.  ANALYSIS.    As  many  yards  as  f  of  a 

12  dollar,  the  price  of  1  yard,  is  contained 

A  times  in  12  dollars.    Integers  cannot  be  di- 

vided by  fourths,  because  they  are  not  of 
3  )  48  the  same  denomination.    Reducing  12  dol- 

16   yards  ^ars  to  fourths  by  multiplying,  we  have  48 

fourths ;  and  3  fourths  is  contained  in  48 
fourths  16  times,  the  required  number  of  yards. 

SECOND  OPERATION.        ANALYSIS.    Here  we  divide  the  integer 
3  \  12  by  the  numerator  of  the  fraction,  and  mul- 

tiply the  quotient  by  the  denominator, 
which  produces  the  same  result  as  in  the 
first  operation.  Hence, 

16   yards. 

Dividing  by  a  fraction  consists  in  multiplying  by  the  denom- 
inator, and  dividing  by  the  numerator  of  the  divisor. 

EXAMPLES   FOR   PRACTICE. 


2.   Divide    18  by  f  . 
3.   Divide    63  by  TV 
4.    Divide    42  by  f 
5.    Divide  120  by  & 
6.   Divide  316  by  ^. 

Ans.     48. 
Ans.   117. 
Ans.     49. 
Ans.   205f 
Ans.  877£. 

Case  II  is  what  ?    Give  first  explanation.     Second.    Deduction. 


DIVISION.  117 

7.  How  many  bushels  of  oats,  worth  f  of  a  dollar  per  bushel, 
will  pay  for  §  of  a  barrel  of  flour,  worth  9  dollars  a  barrel  ? 

Am.    15. 

8.  If  f  of  an  acre  of  land  sell  for  21  dollars,  what  will  an 
acre  sell  for  at  the  same  rate  ?  Am.   $49. 

9.  When  potatoes  are  worth  £  of  a  dollar  a  bushel,  and 
corn  f  of  a  dollar  a  bushel,  how  many  bushels  of  potatoes  are 
equal  in  value  to  16  bushels  of  corn  ?  Am.   22£. 

10.  If  a  man  can  chop  2f  cords  of  wood  in  a  day,  in  how 
many  days  can  he  chop  22  cords  ? 

OPEKATION. 

21  =  -^ 

22  ANALYSIS.    We  reduce  the  mixed  number 

to  an  improper  fraction,  and  then  divide  the 
integer  in  the  same  manner  as  by  a  proper 


11  )88  fraction. 

Am.   8    days. 

11.  Divide  75  by  13f.  Am. 

12.  Divide  149  by  24£.  Am. 

13.  A  farmer  distributed  15  bushels  of  corn  among  some 
poor  persons,  giving  them  If  bushels   apiece;   among  how 
many  persons  did  he  divide  it  ? 

14.  Divide  f  of  320  by  %  of  9£.  Am.   25  f. 

15.  Bought  £  of  7£  cords  of  wood  for  J  of  $32  ;  how  much 
did  1  cord  cost  ?  Am.   $3f 

16.  A  father  divided  183  acres  of  land  equally  among  his 
sons,  giving  them  45  J  acres  apiece  ;  how  many  sons  had  he  ? 

Am.   4. 

CASE   III. 

14£.    To  divide  a  fraction  by  a  fraction. 

1.   How  many  pounds  of  tea  can  be  bought  for  ±£  of  a  dol- 
lar, at  §  of  a  dollar  a  pound  ? 

How  divide  by  a  mixed  number  ?     Case  in  is  what  ? 


118  FRACTIONS. 

OPERATION.  ANALYSIS.      As 

First  step,  f  1  X  3  =  f  £  manY  pounds  as  f 

Second  step,  f  J  -~  2  =  JJ  =  If-  of  a   dollar  is    con- 

11       2*11     "0      11  tained  times  in   11 

Whole  work,  ~  ^  ~     =  ~:X  ~=  —  =lf,^««.of  a  dollar.      1    is 

'12       3       **V  •       8  contained  in  ji  }J 

times,  and  ^  is  con- 

tained in  ||  3  times  as  many  times  as  1,  or  3  times  \%,  which  is  f  f 
times,  which  is  the  number  of  pounds  that  could  be  bought  at  J-  of 
a  dollar  per  pound  ;  but  f  is  contained  but  -|  as  many  times  as  ^, 
and  ||  divided  by  2  gives  ||,  equal  to  1-|  times,  or  the  number  of 
pounds  that  can  be  bought  at  f  of  a  dollar  per  pound. 

We  see  in  the  operation  that  we  have  multiplied  the  dividend  by 
the  denominator  of  the  divisor,  and  divided  the  result  by  the  numer- 
ator of  the  divisor,  which  is  in  accordance  with  14O  for  dividing  a 
fraction.  Hence,  by  inverting  the  terms  of  the  divisor,  the  two 
fractions  will  stand  in  such  relation  to  each  other  that  we  can  mul- 
tiply together  the  two  upper  numbers  for  the  numerator  of  the  quo- 
tient, and  the  two  lower  numbers  for  the  denominator,  as  shown  in 
the  operation.  For  division  of  fractions,  we  have  this  general 

RULE.  I.  Reduce  integers  and  mixed  numbers  to  improper 
fractions. 

II.  Invert  the  terms  of  the  divisor,  and  proceed  as  in  multi- 
plication. 

NOTES.  1.  The  dividend  and  divisor  may  be  reduced  to  a  common 
denominator,  and  the  numerator  of  the  dividend  be  divided  by  the  nu- 
merator of  the  divisor  ;  this  will  give  the  same  result  as  the  rule. 

2.  Apply  cancellation  where  practicable. 

EXAMPLES    FOR   PRACTICE. 


2. 

Divide  |  by  J. 

^t/zs.     1^-. 

3. 

Divide  f  by  £. 

Ans.    3^. 

4. 

Divide  -f-  by  T9jj. 

.te.    t-g. 

5. 

Divide  %  by  /j. 

^4^5. 

G. 

Divide  |  by  f  }. 

Ans.    -jjG. 

7. 

How  many  times  is  f  contained  in  f  ? 

-4ws.  IjV 

8. 

How  many  times  is  y  contained  in  1§  ? 

^tws.    3f. 

Rule,  first  step.     Second.     What  other  method  is  mentioned  ? 


DIVISION.  119 

9.    How  many  times  is  T75  contained  in  ^£  ?      Ans.   2f  . 

10.  How  many  times  is  ^  contained  in  ±%  ? 

11.  How  many  times  is  £  of  £  contained  in  f  of  2£  ? 

12.  What  is  the  quotient  of  T9<y  of  4,  divided  by  £  of  3|  ? 

13.  What  is  the  quotient  of  £  of  I  of  36  divided  by  1J 
times  f  ? 

01 

14.  What  is  the  value  of  —  ? 


OPERATION. 

3£_  J  __7   t  35__#        $  This  example 

-  —  ~^"~^  ~  —  ~7  X  —  ==  f  ,  -4w5.         is  only  another 

4f     -V-     *      8      «     ^5  form    for    ex- 

pressing    divis- 

ion of  fractions  ;  it  is  sometimes  called  a  complex  fraction,  and  the 
process  of  performing  the  division  is  called  reducing  a  complex  frac- 
tion to  a  simple  one. 

We  simply  reduce  the  upper  number  or  dividend  to  an  improper 
fraction,  and  the  lower  number,  or  divisor,  to  an  improper  fraction, 
and  then  divide  as  before. 

15.    What  is  the  value  of  -    ?  Ans.   f  f  . 


16.  What  is  the  value  of  —-?  Ans.    20. 

17.  What  is  the  value  of  -^-P  Ans.   f^ 

4f 

18.  What  is  the  value  of  ?°  *?  Ans.      I. 

* 

P  p        r 

19.  What  is  the  value  of  -     ~  ?  Ans.     £. 


o 

20.  If  a  horse  eat  f  of  a  bushel  of  oats  in  a  day,  in  how 
many  days  will  he  eat  5£  bushels?  Ans.    14. 

21.  If  a  man  spend  If  dollars  per  month  for  tobacco,  in 
what  time  will  he  spend  lOf  dollars?          Ans.    6|  months. 

What  is  a  complex  fraction  ? 


120 


FRACTIONS. 


22.  How  many  times  will  4f  gallons  of  camphene  fill  a 
vessel  that  holds  £  of  |  of  1  gallon  ?  Am.    10J-. 

23.  If  14  acres  of  meadow  land  produce  32  f  tons  of  hay, 
how  many  tons  will  5  acres  produce  ?  Jras.    112. 

24    If  2  yards  of  silk  cost  $3£,  how  much  less  than  $17 
will  9  yards  cost  ?  Ans.    $2|. 

25.  If  f  of  a  yard  of  cloth  cost  ft  of  a  dollar,  how  much 
will  1  yard  cost  ? 

26.  A  man,  having  $10,  gave  §  of  his  money  for  clover 
seed  at  $3£  a  bushel  ;  how  much  did  he  buy?     Ans.    2  bush. 

27.  How  many  tons  of  hay  can  be  purchased  for  $11  9ft, 
at  $9f  per  ton?  Ans. 


PROMISCUOUS    EXAMPLES. 

1.  Reduce  ^,  £,  f,  and  £  to  equivalent  fractions  whose  de- 
nominators shall  be  24.  Ans.   £f  ,  f  £,  2\,  ^. 

2.  Change  i  to  an  equivalent  fraction  having  91  for  its 
denominator  Ans.    ff. 

3.  Find  the  least  common  denominator  of  f,  If,  £  of  f,  2, 
*  of  *  of  1ft. 

4.  Add  4i  f,  $  of  1|,  3,  and  JJ. 

5.  Find  the  difference  between  f  of  6ft  and  f  of  4r\. 

^4»5.     Ifjjf. 

6.  The  less  of  two  numbers  is  475  6  1,  and  their  difference 
is  128f  ;  what  is  the  greater  number?  Ans.    4885/B. 

7.  What  is  the  difference  between  the  continued  products  of 
3,i,S,4f,and3I,S,4,f?  Ans.   3&. 

4         2J- 

8.  Reduce  the  fractions  —  and  —  to  their  simplest  form. 

9.  What  number  multiplied  by  f  will  produce  18257  ? 

Ans.    304.']£. 

10.  A  farmer  had  -£  of  his  sheep  in  one  pasture,  ^  in  an- 
other, and  the  remainder,  which  were  77,  in  a  third  pasture  ; 
how  many  sheep  had  he  ?  Ans.    140. 

11.  What  will  7f  cords  of  wood  cost  at  £  of  9;,  dollars  per 
cord?  Ans.    $24. 


PROMISCUOUS  EXAMPLES.  121 

12.  At  J  of  a  dollar  per  bushel,  how  many  bushels  of  apples 
can  be  bought  for  5|-  dollars? 

13.  Paid  $1837f  for  7350^  bushels  of  oats ;  how  much  was 
that  per  bushel  ?  Ans.   %  of  a  dollar. 

14.  If  235J-  acres  of  land  cost  $4725|,  how  much  will  628 
acres  cost?  Ans.   $12601. 

15.  A  man,  owning  f  of  an  iron  foundery,  sold  £  of  his  share 
for$540£;  what  was  the  value  of  the  foundery  ?  Ans.  $4055 1. 

16.  f  of  |-  of  what  number  is  -         -  less  than  14f  ? 

L4T<J  Ans.    27. 

17.  A  merchant  bought  4£  cords  of  wood  at  $3£  per  cord, 
and  paid  for  it  in  cloth  at  f  of  a  dollar  per  yard  ;  how  many 
yards  were  required  to  pay  for  the  wood  ? 

18.  How  many  yards  of  cloth,  j  of  a  yard  wide,  will  line 
20£  yards,  1^  yards  wide  ?  Ans.   34£. 

19.  If  the  dividend  be  £,  and  the  quotient  ^ff,  what  is  the 
divisor? 

20.  If  the  sum  of  two  fractions  be  |,  and  one  of  them  be 
•/yj  what  is  the  other  ?  Ans.   -/^. 

21.  If  the  smaller  of  two  fractions  be  f  ^,  and  their  differ- 
ence g*j,  what  is  the  greater  ?  Ans.   £f . 

22.  If  3 1  pounds  of  sugar  cost  33  cents,  how  much  must  be 
paid  for  65  £  pounds  ? 

23.  If  324  bushels  of  barley  can  be  had  for  259£  bushels 
of  corn,  how  much  barley  can  be  had  for  2000  bushels  of 
corn?  Ans.    2500  bushels. 

24.  A  certain  sum  of  money  is  to  be  divided  among  5  per- 
sons ;  A  is  to  have  J,  B  £,  C  TV,  D  ^  and  E  the  remainder, 
which  is  20  dollars ;  what  is  the  whole  sum  to  be  divided  ? 

Ans.   $50. 

25.  What  number,  diminished  by  the  difference  between  J 
and  -|  of  itself,  leaves  a  remainder  of  34  ?  Ans.    40. 

26.  If  f  of  a  farm  be  valued  at  $1728,  what  is  the  value  of 
the  whole  ? 


122  FRACTIONS. 

27.  Bought  320  sheep  at  $2£-  per  head ;  afterward  bought 
435  at  $1£  per  head ;  then  sold  f  of  the  whole  number  at  $1J 
per  head,  and  the  remainder  at  $2£;  did  I  gain  or  lose,  and 
how  much  ?  Ans.   Lost  $44£. 

28.  If  5  be  added  to  both  terms  of  the  fraction  £,  will  its 
value  be  increased  or  diminished?        Ans.   Increased  T§¥. 

29.  If  5  be  added  to  both  terms  of  the  fraction  f ,  will  its 
value  be  increased  or  diminished  ?        Ans.  Diminished  ¥5¥. 

30.  How  many  times  can  a  bottle  holding  %  of  §  of  a  gal- 
lon, be  filled  from  a  demijohn  containing  f  of  If  gallons  ? 

Ans.    7£. 

31.  Bought  J  of  7£  cords  of  wood  for  |  of  $32  ;  how  much 
did  1  cord  cost  ? 

32.  Purchased  728  pounds  of  candles  at  1 6|  cents  a  pound ; 
had  they  been  purchased  for  3|-  cents  less  a  pound,  how  many 
pounds  could  have  been  purchased  for  the  same  money  ?  - 

Ans.   953£|. 

33.  What  number,  divided  by  If,  will  give  a  quotient  of 
9£?  Ans.    12  jj. 

34.  The  product  of  two  numbers  is  6,  and  one  of  them  is 
1846;  what  is  the  other?  Ans.    5|5. 

35.  A  stone  mason  worked  11§  days,  and  after  paying  his 
board  and  other  expenses  with  f  of  his  earnings,  he  had  $20 
left ;  how  much  did  he  receive  a  day  ? 

36.  If  f  of  4  tons  of  coal  cost  $5£,  what  will  J  of  2  tons 
cost?  Ans.    $5. 

37.  In  an  orchard  f  of  the  trees  are  apple  trees,  TV  peach 
trees,  and  the  remainder  are  pear  trees,  which  are  20  more  than 
£  of  the  whole ;  how  many  trees  in  the  orchard  ?     Ans.  800. 

38.  A  man  gave  6|  pounds  of  butter,  at  12  cents  a  pound, 
for  £  of  a  gallon  of  oil ;  how  much  was  the  oil  worth  a  gal- 
lon ?  Ans.   $1. 

39.  A  gentleman,  having  271^-  acres  of  land,  sold  £  of  it, 
and  gave  f  of  it  to  his  son ;  what  was  the  value  of  the  re- 
mainder, at  $57|  per  acre  ?  Ans.   $4577B3^. 


PROMISCUOUS   EXAMPLES.  123 

40.  A  horse  and  wagon  cost  $270  ;  the  horse  cost  1£  times 
as  much  as  the  wagon;  what  was  the  cost  of  the  wagon? 

41.  What  number    taken  from  2£  times   12J  will  leave 
20£?  Ans.    11±. 

42.  A  merchant  bought  a  cargo  of  flour  for  $2173^,  and 
sold  it  for  f  §  of  the  cost,  thereby  losing  £  of  a  dollar  per  bar- 
rel;  how  many  barrels  did  he  purchase  ?  Ans.    126. 

43.  A  and  B  can  do  a  piece  of  work  in  14  days ;  A  can  do 
J  as  much  as  B ;  in  how  many  days  can  each  do  it  ? 

Ans.   A,  6  days ;  B,  8  days. 

44.  How  many  yards  of  cloth  f  of  a  yard  wide,  are  equal 
to  12  yards  f  of  a  yard  wide  ?  Ans.    11 J. 

45.  A,  B,  and  C  can  do  a  piece  of  work  in  5  days ;  B  and 
C  can  do  it  in  8  days  ;  in  what  time  can  A  do  it  ? 

46.  A  man  put  his  money  into  4  packages ;  in  the  first  he 
put  f ,  in  th£  second  £,  in  the  third  ^,  and  in  the  fourth  the  re- 
mainder, which  was  $24  more  than  TL  of  the ^v hole ;  how  much 
money  had  he?  Ans.    $720. 

47.  If  $7|-  will  buy  3£  cords  of  wood,  how  many  cords  can 
be  bought  for  $10£  ?  Ans.    4J'. 

48.  How  many  times  is  £  of  f  of  27  contained  in  %  of  ^  of 
42§? 

49.  A  boy  lost  J-  of  his  kite  string,  and  then  added  30  feet, 
when  it  was  just  f  of  its  original  length ;  what  was  the  length 
at  first  ?  Ans.    100  feet. 

50.  Bought  f  of  a  box  of  candles,  and  having  used  £  of 
them,  sold  the  remainder  for  £f  of  a  dollar;  how  much  would 
a  box  cost  at  the  same  rate  ?  Ans.    $5-ff. 

51.  A  post  stands  £  in  the  mud,  £  in  the  water,  and  21  feet 
above  the  water ;  what  is  its  length  ? 

52.  A  father  left  his  eldest  son  f  of  his  estate,  his  youngest 
son  f-  of  the  remainder,  and  his  daughter  the  remainder,  who 
received  $1723f  less  than  the  youngest  son;  what  was  the 
value  of  the  estate  ?  Ans.   $21114£f. 


124  DECIMALS. 


DECIMAL   FRACTIONS. 

Decimal  Fractions  are  fractions  which  have  for 
their  denominator  10,  100,  1000,  or  1  with  any  number  of 
ciphers  annexed. 

NOTES.  1.  The  word  decimal  is  derived  from  the  Latin  decent, 
which  signifies  ten. 

2.  Decimal  fractions  are  commonly  called  decimals. 

3.  Since  ro  z=  iVo '  T¥o" z=  TWo"'  &c->  tne  denominators  of  decimal 
fractions  increase  and  decrease  in  a  tenfold  ratio,  the  same  as  simple 
numbers. 

DECIMAL    NOTATION   AND    NUMERATION. 

144:.  Common  Fractions  are  the  common  divisions  of  a 
unit  into  any  number  of  equal  parts,  as  into  halves,  fifths, 
twenty-fourths,  &c. 

Decimal  Fractions  are  the  decimal  divisions  of  a  unit,  thus : 
A  unit  is  divided  into  ten  equal  parts,  called  tenths ;  each  of 
these  tenths  is  divided  into  ten  other  equal  parts  called  hun- 
dredths ;  each  of  these  hundredths  into  ten  other  equal  parts, 
called  thousandths;  and  so  on.  Since  the  denominators  of 
decimal  fractions  increase  and  decrease  by  the  scale  of  10,  the 
same  as  simple  numbers,  in  writing  decimals  the  denomina- 
tors may  be  omitted. 

In  simple  numbers,  the  unit,  1,  is  the  starting  point  of 
notation  and  numeration ;  and  so  also  is  it  in  decimals.  We 
extend  the  scale  of  notation  to  the  left  of  units'  place  in 
writing  integers,  and  to  the  right  of  units'  place  in  writing 
decimals.  Thus,  the  first  place  at  the  left  of  units  is  tens, 
and  the  first  place  at  the  right  of  units  is  tenths  ;  the  second 
place  at  the  left  is  hundreds,  and  the  second  place  at  the 
right  is  hundredths  ;  the  third  place  at  the  left  is  thousands, 
and  the  third  place  at  the  right  is  thousandths ;  and  so  on. 


What  are  decimal  fractions?     How  do  they  differ  from  common 
fractions  ?     How  are  they  written  ? 


NOTATION  AND  NUMERATION.          125 

The  Decimal  Point  is  a  period  (  .  ),  which  must  always  be 
placed  before  or  at  the  left  hand  of  the  decimal.     Thus, 
T6u    is  expressed  .6 

"  '54 

«          .279 


NOTE.  The  decimal  point  is  also  called  the  Separatrix.  This  is  a 
correct  name  for  it  only  when  it  stands  between  the  integral  and  deci- 
mal parts  of  the  same  number. 

.5       is  5  tenths,  which    =  -^  of  5  units  ; 

.05     is  5  hundredths,  "       =  TV  of  5  tenths  ; 

.005  is  5  thousandths,          "       =  T\j  of  5  hundredths. 

And  universally,  the  value  of  a  figure  in  any  decimal  place 
is  y1^  the  value  of  the  same  figure  in  the  next  left  hand  place. 

The  relation  of  decimals  and  integers  to  each  other  is  clear- 
ly shown  by  the  following 

NUMERATION  TABLE. 


5 
•s.3  « 

SlS 


4753.62418695 

Integers.  Decimals. 

By  examining  this  table  we  see  that 

Tenths  are  expressed  by  one    figure. 

Hundredths  "  "          "  two     figures. 

Thousandths         "  "          "  three       « 

Ten  thousandths  «  "          "  four         " 

And  any  order  of  decimals  by  one  figure  less  than  the  corre- 
sponding order  of  integers. 

14<»>.    Since  the   denominator  of  tenths  is    10,  of  hun- 

What  is  the  decimal  point  ?     What  is  it  sometimes  called  ?     "What  is 
the  value  of  a  figure  in  any  decimal  place  ? 

K* 


126  DECIMALS. 

dredths  100,  of  thousands  1000,  and  so  on,  a  decimal  may  be 
expressed  by  writing  the  numerator  only  ;  but  in  this  case 
the  numerator  or  decimal  must  always  contain  as  many 
decimal  places  as  are  equal  to  the  number  of  ciphers  in  the 
denominator  ;  and  the  denominator  of  a  decimal  will  always 
be  the  unit,  1,  with  as  many  ciphers  annexed  as  are  equal  to 
the  number  of  figures  in  the  decimal  or  numerator. 
The  decimal  point  must  never  be  omitted. 

EXAMPLES    FOR    PRACTICE. 

1.  Express  in  figures  thirty-eight  hundredths. 

2.  Write  seven  tenths. 

3.  Write  three  hundred  twenty-five  thousandths. 

4.  Write  four  hundredths.  Ans.   .04. 

5.  Write  sixteen  thousandths. 

6.  Write  seventy-four  hundred-thousandths.    Ans.  .00074. 

7.  Write  seven  hundred  forty-five  millionths. 

8.  Write  four  thousand  two  hundred  thirty-two  ten-thou- 
sandths. 

9.  Write  five  hundred  thousand  millionths. 
10.    Read  the  following  decimals  : 

.05  .681  .9034  .19248 

.24  .024  .0005  .001385 

.672  .8471  .100248  .1000087 

NOTE.  To  read  a  decimal,  we  first  numerate  from  left  to  right,  and 
the  name  of  the  right  hand  figure  is  the  name  of  the  denominator.  We 
then  numerate  from  right  to  left,  as  in  whole  numbers,  to  read  the 
numerator. 


A  mixed  number  is  a  number  consisting  of  integers 
and  decimals;  thus,  71.406  consists  of  the  integral  part,  71, 
and  the  decimal  part,  .406;  it  is  read  the  same  as  7 
71  and  406  thousandths. 


EXAMPLES    FOR    PRACTICE. 


1.  Write  eighteen,  and  twenty-seven  thousandths. 

2.  Write  four  hundred,  and  nineteen  ten-millionths. 


How  many  decimal  places  must  there  be  to  express  any  decimal  ? 


KOTATION   AND   NUMERATION.  127 

3.  Write  fifty-four,  and  fifty-four  millionths. 

4.  Eighty-one,  and  1  ten-thousandth. 

5.  One  hundred,  and  67  ten-thousandths. 

6.  Read  the  following  numbers : 

18.027  100.0067  400.0000019 

81.0001  54.000054  3.03 

75.075  9.2806  40.40404 

147.  From  the  foregoing  explanations  and  illustrations 
we  derive  the  following  important 

PRINCIPLES    OF   DECIMAL    NOTATION   AND    NUMERATION. 

1.  The  value  of  any  decimal  figure  depends  upon  its  place 
from  the  decimal  point :  thus  .3  is  ten  times  .03. 

2.  Prefixing  a  cipher  to  a  decimal  decreases  its  value  the 
same  as  dividing  it  by  ten ;  thus,  .03  is  T^  the  value  of  .3. 

3.  Annexing  a  cipher  to  a  decimal  does  not  alter  its  value, 
since  it  does  not  change  the  place  of  the  significant  figures  of 
the  decimal ;  thus,  T6ff,  or  .6,  is  the  same  as  T6n%  or  .60. 

4.  Decimals  increase  from  right  to  left,  and  decrease  from 
left  to  right,  in  a  tenfold  ratio ;  and  therefore  they  may  be 
added,  subtracted,  multiplied,  and  divided  the  same  as  whole 
numbers. 

5.  The  denominator  of  a  decimal,  though  never  expressed, 
is  always  the  unit,  1,  with  as  many  ciphers  annexed  as  there 
are  figures  in  the  decimal. 

6.  To  read  decimals  requires  two  numerations  ;  first,  from 
units,  to  find  the  name  of  the  denominator,  and  second,  towards 
units,  to  find  the  value  of  the  numerator. 

148.  Having  analyzed  all  the  principles  upon  which  the 
writing  and  reading  of  decimals  depend,  we  will  now  present 
these  principles  in  the  form  of  rules. 

RULE    FOR   DECIMAL    NOTATION. 

I.     Write  the  decimal  the  same  as  a  whole  number,  placing 

"What  is  the  first  principle  of  decimal  notation  ?     Second  ?     Third  ? 
Fourth  ?     Fifth  ?     Sixth  ?     Kule  for  notation,  first  step  ? 


128  DECIMALS. 

ciphers  where  necessary  to  give  each  significant  figure  its  true 
local  value. 

II.   Place  the  decimal  point  before  the  first  figure. 

RULE   FOR   DECIMAL   NUMERATION. 

I.  Numerate  from  the  decimal  point 9  to  determine  the  de- 
nominator. 

II.  Numerate  towards  the  decimal  pointy  to  determine  the 
numerator. 

III.  Read  the  decimal  as  a  whole  number,  giving  it  the  name 
or  denomination  of  the  right  hand  figure. 

EXAMPLES    FOR   PRACTICE. 

1.  Write  425  million ths. 

2.  Write  six  thousand  ten-thousandths. 

3.  Write  one  thousand  eight  hundred  fifty-nine  hundred- 
thousandths. 

4.  Write  260  thousand  8  billionths. 

5.  Read  the  following  decimals : 

.6321  .748243  .2962999 

.5400027  .60000000  .00000006 

6.  Write  five  hundred  two,  and  one  thousand  six  millionths. 

7.  Write  thirty-one,  and  two  ten-millionths. 

8.  Write  eleven  thousand,  and  eleven  hundred-thousandths. 

9.  Write  nine  million,  and  nine  billionths. 

10.  Write  one  hundred  two  tenths.  Am.    10.2. 

11.  Write  one  hundred  twenty-four  thousand  three   hun- 
dred fifteen  thousandths. 

12.  Write  seven  hundred  thousandths. 

13.  Write  seven  hundred-thousandths. 

14.  Read  the  following  numbers : 

12.36  9.052  62.9999 

142.847  32.004  1858.4583 

1.02  4.0005  27.00045 

Second  ?    Rule  for  numeration,  first  step  ?     Second  ?    Third  ? 


REDUCTION.  129 


REDUCTION. 
CASE  I. 

149.  To  reduce  decimals  to  a  common  denomina- 
tor. 

1.  Reduce  .5,  .375,  3.25401,  and  46.13  to  their  least  com- 
mon decimal  denominator. 

OPEEATION.  ANALYSIS.    A  common  denominator  must  con- 

.50000         ta^n  as  many  decimal  places  as  is  equal  to  the 

37500          greatest  number  of  decimal  figures  in  any  of  the 

3*25401          given  decimals.     We  find  that  the  third  number 

'  contains  five  decimal  places,  and  hence  100000 

"  must  be  a  common  denominator.     As  annexing 

ciphers  to  decimals  does  not  alter  their  value,(l  44:., 3)we  give  to  each 

number  five  decimal  places  by  annexing  ciphers,  and  thus  reduce  the 

given  decimals  to  a  common  denominator.     Hence, 

RULE.  Give  to  each  number  the  same  number  of  decimal 
places,  by  annexing  ciphers. 

NOTES.  1.  If  the  numbers  be  reduced  to  the  denominator  of  that 
one  of  the  given  numbers  having  the  greatest  number  of  decimal  places, 
they  will  have  their  least  common  decimal  denominator. 

2.  A  whole  number  may  readily  be  reduced  to  decimals  by  placing 
the  decimal  point  after  units,  and  annexing  ciphers ;  one  cipher  re- 
ducing it  to  tenths,  two  ciphers  to  hundredths,  three  ciphers  to  thou- 
sandths, and  so  on. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  .17,  24.6,  .0003,  84,  and  721.8000271  to  their 
least  common  denominator. 

3.  Reduce  7  tenths,  24  thousandths,  187  millionths,  5  hun- 
dred millionths,  and  10845  hundredths  to  their  least  common 
denominator. 

4.  Reduce  to  their  least  common  denominator  the  following 
decimals:  1000.001,  841.78,  2.6004,  90.000009,  and  6000. 

What  is  meant  by  the  reduction  of  decimals  ?  Case  I  is  what  ? 
Give  explanation.  Rule. 


130  DECIMALS. 

CASE    II. 

150.  To  reduce  a  decimal  to  a  common  fraction. 

1.  Reduce  .75  to  its  equivalent  common  fraction. 

OPERATION.  ANALYSIS.     We  omit  the  decimal  point, 

^75  _.  _7  5    —  a.          supply  the  proper  denominator  to  the  deci- 
mal, and  then  reduce  the  common  fraction 
thus  formed  to  its  lowest  terms.     Hence, 

RULE.     Omit   the  decimal  point,  and  supply  the  proper 
denominator. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  .125  to  a  common  fraction.  Ans.  £. 

3.  Reduce  .16  to  a  common  fraction.  Ans.  -/$. 

4.  Reduce  .655  to  a  common  fraction.  Ans.  £$^. 

5.  Reduce  .9375  to  a  common  fraction.  Ans.  -ff . 

6.  Reduce  .0008  to  a  common  fraction.  Ans.  T2Ttf. 

CASE    III. 

151.  To  reduce  a  common  fraction  to  a  decimal. 

1.  Reduce  f-  to  its  equivalent  decimal. 

FIRST  OPERATION.  ANALYSIS.    We  first  annex 

f.  =  J $$  =  TVfr  =  .75,  Ans.         tlie  same  number  of  ciphers 

to  both  terms  of  the  fraction ; 

SECOND  OPERATION.  this  d°es  not  alter  its  value. 

»  \  o  f\f)  We  then  divide  both  resulting 

terms  by  4,  the  significant  fig- 
.75  ure  of  the  denominator,  to  ob- 

tain the  decimal  denominator, 

100.  Then  the  fraction  is  changed  to  tEe  decimal  form  by  omitting 
the  denominator.  If  the  intermediate  steps  be  omitted,  the  true 
result  may  be  obtained  as  in  the  second  operation. 

2.  Reduce  TX6  to  its  equivalent  decimal. 

Case  II  is  what?     Give  explanation.     Rule.      Case  III  is  what? 
Explain  first  operation.     Second. 


REDUCTION.  131 

THIRD  OPERATION.  ANALYSIS.     Dividing  as  in  the  former 

16)  1.0000  example,  we  obtain  a  quotient  of  3  fig- 

ures,    625.      But    since   we   annexed   4 

.Ub2o,  Ans.         ciphers,  there  must  be  4  places  in  the 

required  decimal;   hence  we  prefix  1  cipher.     This  is  made  still 

plainer  by  the  following  operation ;  thus, 

A  =  iWftfr  =  T*fcfo  =  -0625. 
From  these  illustrations  we  derive  the  following 

RULE.  I.  Annex  ciphers  to  the  numerator,  and  divide  by 
the  denominator. 

II.  Point  off  as  many  decimal  places  in  the  result  as  are 
equal  to  the  number  of  ciphers  annexed. 

NOTE.  A  common  fraction  can  be  reduced  to  an  exact  decimal  when 
its  lowest  denominator  contains  only  the  prime  factors  2  and  5,  and 
not  otherwise. 

EXAMPLES    FOR   PRACTICE. 

3.  Reduce  £  to  a  decimal.  Ans.  .625. 

4.  Reduce  f  to  a  decimal. 

5.  Reduce  J-f  to  a  decimal.  Ans.          .9375. 

6.  Reduce  J  to  a  decimal. 

7.  Reduce  ^  to  a  decimal.  Ans.  .08. 

8.  Reduce  -fy  to  a  decimal.  Ans.      .046875. 

9.  Reduce  f  to  a  decimal. 

10.  Reduce  F3^  to  a  decimal. 

11.  Reduce  Bg^  to  a  decimal.  Ans.  .00375. 

12.  Reduce  T£F  to  a  decimal.  Ans.  .008. 

13.  Reduce  £  to  a  decimal.  Ans.  .33333+. 

NOTE.  The  sign,  -f-,  in  the  answer  indicates  that  there  is  still  a 
remainder. 

14.  Reduce  £f  to  a  decimal.  Ans.   .513513+. 

NOTE.  The  answers  to  the  last  two  examples  are  called  repeating 
decimals ;  and  the  figure  3  in  the  13th  example,  and  the  figures  513  in 
the  14th,  are  called  repetends,  because  they  are  repeated,  or  occur  in 
regular  order. 


Third  operation.     Rule,  first  step  ?     Second  ?     When  can  a  common 
fraction  be  reduced  to  an  exact  decimal  ? 


132 


DECIMALS. 


ADDITION. 


20.074  ? 


1.   What  Is  the  sum  of  3.703,  621.57,  .672,  and 


OPERATION. 
3.703 

621.57 

.672 
20.0074 

645.9524 


ANALYSIS.  We  write  the  numbers  so  that  fig- 
ures of  like  orders  of  units  shall  stand  in  the  same 
columns ;  that  is,  units  under  units,  tenths  under 
tenths,  hundredths  under  hundredths,  &c.  This 
brings  the  decimal  points  directly  under  each 
other.  Commencing  at  the  right  hand,  we  add 
each  column  separately,  and  carry  as  in  whole 
numbers,  and  in  the  result  we  place  a  decimal 
point  between  units  and  tenths,  or  directly  under  the  decimal  point 
in  the  numbers  added.  From  this  example  we  derive  the  following 

RULE.  I.  Write  the  numbers  so  that  the  decimal  points 
shall  stand  directly  under  each  other. 

II.  Add  as  in  ivhole  numbers,  and  place  the  decimal  point, 
in  the  result,  directly  under  the  points  in  the  numbers  added. 


EXAMPLES    FOR    PRACTICE. 


2.  Add 


Sum, 


.199 

2.7569 

.25 

.654 

3.8599 


3.  Add 


4.015 
6.75 

27.38203 
375.01 
2.5 


Amount,     415.65703 

4.  Add  1152.01,  14.11018,  152348.21,  9.000083. 

Am.    153523.330263. 

5.  Add  37.03,  0.521,  .9,  1000,  4000.0004. 

Ans.   5038.4514. 

6.  What  is  the  sum  of  twenty-six,  and   twenty-six  hun- 
dredths ;    seven  tenths ;    six,  and   eighty-three   thousandths ; 
four,  and  four  thousandths  ?  Ans.    37.047. 


Explain  the  operation  of  addition  of  decimals.     Give  rule,  first  step. 
Second. 


ADDITION.  133 

7.  What  is  the  sum  of  thirty-six,  and  fifteen  thousandths  ; 
three    hundred,  and  six  hundred  five  ten-thousandths ;    five, 
and  three  millionths  ;  sixty,  and  eighty-seven  ten-millionths  ? 

Ans.   401.0755117. 

8.  "What  is    the    sura    of   fifty-four,  and  thirty-four  hun- 
dredths  ;  one,  and  nine  ten-thousandths  ;  three,  and  two  hun- 
dred seven  millionths;  twenty-three  thousandths;  eight,  and 
nine  tenths ;  four,  and  one  hundred  thirty-five  thousandths  ? 

Ans.   71.399107. 

9.  How  many  yards  in  three  pieces  of  cloth,  the  first  piece 
containing  18.375  yards,  the  second  piece  41.625  yards,  and 
the  third  piece  35.5  yards? 

10.  A's  farm   contains  61.843  acres,  B's  contains  143.75 
acres,  C's  218.4375   acres,  and  D's    21.9  acres;  how  many 
acres  in  the  four  farms  ? 

11.  My  farm  consists  of  7  fields,  containing  12}  acres,  18f 
acres,  9  acres,  24|  acres,  4||  acres,  8^  acres,  and  15^§  acres 
respectively ;  how  many  acres  in  my  farm  ? 

NOTE.     Reduce  the  common  fractions  to  decimals  before  adding. 

Ans.   93.6375. 

12.  A  grocer  has  2£  barrels  of  A  sugar,  5|  barrels  of  B 
sugar,    3|    barrels   of   C  sugar,    3.0642    barrels    of  crushed 
sugar,  and  8.925  barrels  of  pulverized  sugar ;  how  many  bar- 
rels of  sugar  has  he  ?  Ans.   23.8642. 

13.  A  tailor  made  3  suits  of  clothes ;  for  the  first  suit  he 
used  2£  yards  of  broadcloth,  3^  yards  of  cassimere,  and  £ 
yards  of  satin  ;  for  the  second  suit  2.25  yards  of  broadcloth, 
2.875  yards  of  cassimere,  and  1  yard  of  satin ;  and  for  the 
third    suit  5-^  yards  of  broadcloth,  and  1£   yards   of  satin. 
How  many  yards  of  each  kind  of  goods  did  he  use  ?     How 
many  yards  of  all  ?  Ans.  to  last,   18.375. 

L 


134  DECIMALS. 


SUBTRACTION. 

153.    1.   From  91.73  take  2.18.    ANALYSIS.  In  each  of  these 

three  examples,  we  write  the 

OPERATION,  subtrahend  under  the  minu- 

91-73  end,    placing    units    under 

2.18  units,  tenths  under  tenths, 

Ans.   89.55  ^c*     Commencing   at    the 

right  hand,  we  subtract  as 

2.  From  2.9185  take  1.42.  *?  wholc  ™mbers>  ™d  f 

the  remainders  we  place  the 

OPERATION.  decimal  points  directly  under 

2.9185  those  in  the  numbers  above. 

j  ^2  In  the  second  example,  the 

number  of  decimal  places  in 

Ans.    1.4985  the  mmuend  is  greater  than 

the  number  in  the  subtra- 

3.  From  12465  take  95.58746.     hend]  and  in  the  third  exam. 

OPERATION.  pie  the  number  is  less.     In 

124.65  koth  cases,  we  reduce  both 

or  "OVA a  mmuend  and  subtrahend  to 

VO.tJOl  1O  .  .         , 

the  same  number  of  decimal 

Ans.    29.06254  places,  by  annexing  ciphers; 

or  we  suppose  the  ciphers  to 
be  annexed,  before  performing  the  subtraction.     Hence  the 

RULE.  I.  Write  the  numbers  so  that  the  decimal  points 
shall  stand  directly  under  each  other, 

II.  Subtract  as  in  whole  numbers,  and  place  the  decimal 
point  in  the  result  directly  under  the  points  in  the  given  numbers. 

4.  Find  the  difference  between  714  and  .91 6.    Ans.  713.084. 

5.  How  much  greater  is  2  than  .298  ?  Ans.    1.702. 

6.  From  21.004  take  75  hundredths. 

7.  From  10.0302  take  2  ten-thousandths.      Ans.    10.03. 

8.  From  900  take  .009.  Ans.    899.991. 

9.  From  two  thousand  take  two  thousandths. 

10.    From  one  take  one  millionth.  Ans.    .999999. 


Explain  subtraction  of  fractions.     Give  the  rule,  first  step.     Second. 


MULTIPLICATION.  135 

11.  From   four   hundred   twenty-seven    thousandths    take 
four  hundred  twenty-seven  millionths.  Ans.   .426573. 

12.  A  man  owned  thirty-four  hundredths  of  a  township  of 
land,  and  sold  thirty-four  thousandths  of  the  township ;  how 
much  did  he  still  own  ?  Ans.   .306. 


MULTIPLICATION. 

154.    1.   What  is  the  product  of  .35  multiplied  by  .5  ? 

OPERATION.         ANALYSIS.     We  perform  the  multiplication  the 

.35  same  as  in  whole  numbers,  and  the  only  difficulty 

5  we  meet  with  is  in  pointing  off  the  decimal  places 

in  the  product.     To  determine  how  many  places  to 

.175,  Ans.  p0int  Off,  we  may  reduce  the  decimals  to  common 
fractions ;  thus,  .35  n=  T3^  and  .5  =.  T6^.  Perform- 
ing the  multiplication,  and  we  have  ^  X  -&  =  ^jfe,  and  this 
product,  expressed  decimally,  is  .175.  Here  we  see  that  the  prod- 
uct contains  as  many  decimal  places  as  are  contained  in  both  mul- 
tiplicand and  multiplier.  Hence  the  following 

RULE.  Multiply  as  in  whole  numbers,  and  from  the  right 
hand  of  the  product  point  off  as  many  figures  for  decimals  as 
there  are  decimal  places  in  both  factors. 

NOTES.  1.  If  there  be  not  as  many  figures  in  the  product  as  there 
are  decimals  in  both  factors,  supply  the  deficiency  by  prefixing  ciphers. 

2.  To  multiply  a  decimal  by  10,  100,  1000,  &c.,  remove  the  point  as 
many  places  to  the  right  as  there  are  ciphers  on  the  right  of  the  multi- 
plier." 

EXAMPLES. 

2.  Multiply  1.245  by  .27.  Ans.   .33615. 

3.  Multiply  79.347  by  23.15.  Ans.    1836.88305. 

4.  Multiply  350  by  .7853. 

5.  Multiply  one  tenth  by  one  tenth.  Ans.   .01. 

6.  Multiply  25  by  twenty-five  hundredths.     Ans.    6.25. 


Explain  multiplication  of  decimals.  Give  rule.  If  the  product  have 
less  decimal  places  than  both  factors,  how  proceed  ?  How  multiply  by 
10,  100,  1000,  &c.  ? 


136  DECIMALS. 

7.  Multiply  .132  by  .241.  Ans.  .031812. 

8.  Multiply  24.35  by  10. 

9.  Multiply  .006  by  1000.  Ans.  6. 

10.  Multiply  .23  by  .009.  Ans.    .00207. 

11.  Multiply  sixty-four  thousandths  by  thirteen  million ths. 

Ans.   .000000832. 

12.  Multiply  eighty-seven  ten-thousandths  by  three  hun- 
dred fifty-two  hundred-thousandths. 

13.  Multiply  one  million  by  one  millionth.  Ans.    1. 

14.  Multiply  sixteen  thousand  by  sixteen  ten-thousandths. 

Ans.   25.6. 

15.  If  a  cord  of  wood  be  worth  2.37  bushels  of  wheat,  how 
many  bushels  of  wheat  must  be  given  for  9.58  cords  of  wood  ? 

Ans.   22.7046  bushels. 


DIVISION. 

155.    1.   What  is  the  quotient  of  .175  divided  by  .5  ? 

OPERATION.       ANALYSIS.    We  perform  the  division  the  same  as 
.5  )  .175       in  whole  numbers,  and  the  only  difficulty  we  meet 
-       with  is  in  pointing  off  the  decimal  places  in  the  quo- 
Ans.   «35          tient.     To  determine  how  many  places  to  point  off, 
we  may  reduce  the  decimals  to  common  fractions;  thus,  .175  = 
an(^  -5  —  iV     Performing  the  division,  and  we  have 


175       5       M$      10       35 

VX         _  -•.!.• 


1000      10      1000      £        100 

and  this  quotient,  expressed  decimally,  is  .35.  Here  we  see  that  the 
dividend  contains  as  many  decimal  places  as  are  contained  in  both 
divisor  and  quotient.  Hence  the  following 

RULE.  Divide  as  in  whole  numbers,  and  from  the  right 
hand  of  the  quotient  point  off  as  many  places  for  decimals 
as  the  decimal  places  in  the  dividend  exceed  those  in  the 
divisor. 

Explain  division  of  decimals.    Give  rule. 


DIVISION.  137 

NOTES.  1.  If  the  number  of  figures  in  the  quotient  be  less  than  the 
excess  of  the  decimal  places  in  the  dividend  over  those  in  the  divisor, 
the  deficiency  must  be  supplied  by  prefixing  ciphers. 

2.  If  there  be  a  remainder  after  dividing  the  dividend,  annex  ciphers, 
and  continue  the  division:  the  ciphers  annexed  are  decimals  of  the 
dividend. 

3.  The  dividend  must  always  contain  at  least  as  many  decimal  places 
as  the  divisor,  before  commencing  the  division. 

4.  In  most  business  transactions,  the  division  is  considered  suffi- 
ciently exact  when  the  quotient  is  carried  to  4  decimal  places,  unless 
great  accuracy  is  required. 

5.  To  divide  by  10,    100,  1000,  &c.,  remove  the  decimal  point  as 
many  places  to  the  left  as  there  are  ciphers  on  the  right  hand  of  the 
divisor. 


EXAMPLES   FOR   PRACTICE. 

2.  Divide  .675  by  .15.  Am.       4.5. 

3.  Divide  .288  by  3.6.  Ans.       .08. 

4.  Divide  81.6  by  2.5.  Ans.  32.64. 

5.  Divide  2.3421  by  21.1. 

6.  Divide  2.3421  by  .211. 

7.  Divide  8.297496  by  .153.  Ans.       54.232. 

8.  Divide  12  by  .7854. 

9.  Divide  3  by  3  ;  divide  3  by  .3 ;  3  by  .03 ;  30  by  .03. 

10.  Divide  15.34  by  2.7. 

11.  Divide  .1  by  .7.  Ans.   .142857+. 

12.  Divide  45.30  by  .015.  Ans.  3020. 

13.  Divide  .003753  by  625.5.  Ans.        .000006. 

14.  Divide  9.  by  450.  Ans.  .02. 

15.  Divide  2.39015  by  .007.  Ans.         341.45. 

16.  Divide  fifteen,  and  eight  hundred  seventy-five  thou- 
sandths,by  twenty-five  ten-thousandths.  Ans.    6350. 

17.  Divide  365  by  100. 

18.  Divide  785.4  by  1000.  Ans.   .7854. 

19.  Divide  one  thousand  by  one  thousandth. 

Ans.   1000000. 


When  are  ciphers  prefixed  to  the  quotient  ?  If  there  be  a  remainder, 
how  proceed  ?  If  the  dividend  have  less  decimal  places  than  the  divi- 
sor, how  proceed  ?  How  divide  by  10,  100,  1000,  &c.  ? 

L* 


138  DECIMALS. 

PROMISCUOUS     EXAMPLES. 

1.  Add  six   hundred,  and   twenty  -five   thousandths  ;    four 
tenths  ;  seven,  and  sixty-two  ten-thousandths  ;  three,  and  fifty- 
eight  millionths  ;  ninety-two,  and  seven  hundredths. 

Ans.  702.501258. 

2.  What    is  the    sum    of    81.003  +  5000.4+5.0008  + 
73.87563  +  1000  +  25  +  3.000548  +  .0315  ? 

3.  From  eighty-seven  take  eighty-seven  thousandths. 

4.  What  is  the  difference  between  nine  million  and  nine 
millionths?  Ans.   8999999.999991. 

5.  Multiply  .365  by  .15.  Ans.   .05475. 

6.  Multiply  three  thousandths  by  four  hundredths. 

7.  If  one  acre  produce  42.57  bushels  of  corn,  how  many 
bushels  will  18.73  acres  produce  ?  Ans.   797.3361. 

8.  Divide  .125  by  8000.  Ans.   .000015625. 

9.  Divide  .7744  by  .1936. 

10.  Divide  27.1  by  100000.  Ans.   .000271. 

11.  If  6.35  acres  produce  70.6755  bushels  of  wheat,  what 
does  one  acre  produce  ?  Ans.    11.13  bushels. 

12.  Reduce  .625  to  a  common  fraction.  Ans.   f. 

13.  Express  26.875  by  an  integer  and  a  common  fraction. 

Ans.   26£. 

14.  Reduce  Tf?  to  a  decimal  fraction.  Ans.  .016. 

15.  Reduce         to  a  decimal  fraction.  Ans.  .5. 


16.  How  many  times  will  .5  of  1.75  be  contained  in  .25  of 
17£?  Ans.  5. 

17.  What  will  be  the  cost  of  3|  bales  of  cloth,  each  bale 
containing  36.75  yards,  at  .85  dollars  per  yard  ? 

18.  Traveling  at  the  rate  of  4f  miles  an  hour,  how  many 
hour*  will  a  man  require  to  travel  56.925  miles. 

Ans.    12$  hours. 


NOTATION   AND  NUMERATION.  139 


DECIMAL   CURRENCY. 

156.  Coin  is  money  stamped,  and  has  a  given  value  es- 
tablished by  law. 

157.  Currency  is  coin,  bank  bills,  treasury  notes,  &c.,  in 
circulation  as  a  medium  of  trade. 

158 .  A  Decimal  Currency  is  a  currency  whose  denom- 
inations increase  and  decrease  in  a  tenfold  ratio. 

NOTE.  The  currency  of  the  United  States  is  decimal  currency,  and 
is  sometimes  called  Federal  Money ,•  it  was  adopted  by  Congress  in  1786. 

NOTATION    AND    NUMERATION. 

The  gold  coins  of  the  United  States  are  the  double  eagle, 
half  eagle,  and  quarter  eagle,  three  dollar  piece,  and  dollar. 

The  silver  coins  are  the  dollar,  half  and  quarter  dollar,  dime 
and  half  dime,  and  three  cent  piece. 

The  nickel  coin  is  the  cent. 

NOTES.  1.  The  following  pieces  of  gold  are  in  use,  but  are  not  legal 
coin,  viz. ;  the  fifty  dollar  piece,  and  the  half  and  quarter  dollar  pieces. 

2.  The  copper  cent  and  half  cent,  though  still  in  circulation,  are  no 
longer  coined. 

3.  The  mill  is  used  only  in  computation ;  it  is  not  a  coin. 

TABLE. 

10  mills  (m.)  make  1  cent,     .  .  .  c. 

10  cents  "       1  dime,    .  .  .  d. 

10  dimes  "       1  dollar,  .  .  .  $. 

10  dollars  "       1  eagle,  .  .  .  E. 

UNIT    EQUIVALENTS. 
Mills.  Cents. 

10  =    1  Dimes. 

100         =10         =    1  Dollarg. 

1000      =    100      =10      =    I  Eagle. 

10000  =  1000  =  100  =  10  =  1 

NOTE.  The  character  $  is  supposed  to  be  a  contraction  of  U.  S., 
(United  States,")  the  U  being  placed  upon  the  S. 

"What  is  coin?  Currency?  Decimal  currency?  Federal  money? 
What  are  the  gold  coins  of  U.  S.  ?  Silver  ?  Copper  ?  What  are  the 
denominations  of  U.  S.  currency  ?  What  is  the  sign  of  dollars  ?  From 
what  derived  ? 


140  DECIMAL   CURRENCY. 

159*  The  dollar  is  the  unit  of  United  States  money; 
dimes,  cents,  and  mills  are  fractions  of  a  dollar,  and  are  sepa- 
rated from  the  dollar  by  the  decimal  point ;  thus,  two  dollars 
one  dime  two  cents  five  mills,  are  written  $2.125. 

By  examining  the  table,  we  see  that  the  dime  is  a  tenth  part 
of  the  unit,  or  dollar ;  the  cent  a  tenth  part  of  the  dime  or  a 
hundredth  part  of  the  dollar ;  and  the  mill  a  tenth  part  of  the 
cent,  a  hundredth  part  of  the  dime,  or  a  thousandth  part  of  the 
dollar.  Hence  the  denominations  of  decimal  currency  increase 
and  decrease  the  same  as  decimal  fractions,  and  are  expressed 
according  to  the  same  decimal  system  of  notation ;  and  they 
may  be  added,  subtracted,  multiplied,  and  divided  in  the  same 
manner  as  decimals. 

Dimes  are  not  read  as  dimes,  but  the  two  places  of  dimes 
and  cents  are  appropriated  to  cents ;  thus,  1  dollar  3  dimes 
2  cents,  or  $1.32,  are  read  one  dollar  thirty-two  cents  ;  hence, 

When  the  number  of  cents  is  less  than  1 0,  we  write  a  cipher 
before  it  in  the  place  of  dimes. 

NOTE.  The  half  cent  is  frequently  written  as  5  mills  ;  thus,  24<|  cents, 
written  $.245. 

1OO.  Business  men  frequently  write  cents  as  common 
fractions  of  a  dollar  ;  thus,  three  dollars  thirteen  cents  are 
written  SSy1^,  and  read,  three  and  thirteen  hundredths  dollars. 
In  business  transactions,  when  the  final  result  of  a  computation 
contains  5  mills  or  more,  they  are  called  one  cent,  and  when 
less  than  5,  they  are  rejected. 

EXAMPLES   FOR   PRACTICE. 

1.  Write  four  dollars  five  cents.  Ans.   $4.05. 

2.  Write  two  dollars  nine  cents. 

3.  Write  ten  dollars  ten  cents. 

4.  Write  eight  dollars  seven  mills.  Ans.    $8.007. 

What  is  the  unit  of  U.  S.  currency  ?  What  is  the  general  law  of 
increase  and  decrease  ?  In  practice,  how  many  decimal  places  are  given 
to  cents  ?  In  business  transactions,  how  are  cents  frequently  written  ? 
What  is  done  if  the  mills  exceed  5  ?  If  less  than  5  ? 


REDUCTION.  141 

5.  Write  sixty-four  cents.  Ans.   $0.64. 

6.  Write  three  cents  two  mills. 

7.  Write  one  hundred  dollars  one  cent  one  mill. 

8.  Read  $7.93 ;  $8.02  ;  $6.542. 

9.  Read  $5.272;  $100.025;  $17.005. 

10.   Read  $16.205;  $215.081;  $1000.011;  $4.002. 

REDUCTION. 

161.  By  examining  the  table  of  Decimal  Currency,  we  see 
that  10  mills  make  one  cent,  and  100  cents,  or  1000  mills, 
make  one  dollar ;  hence, 

To  change  dollars  to  cents,  multiply  by  100 ;  that  is,  annex 
two  ciphers. 

To  change  dollars  to  mills,  annex  three  ciphers. 
To  change  cents  to  mills,  annex  one  cipher. 

EXAMPLES    FOR   PRACTICE. 

1.  Change  $792  to  cents.  Ans.   79200  cents. 

2.  Change  $36  to  cents. 

3.  Reduce  $5248  to  cents. 

4.  In  6.25  dollars  how  many  cents  ?        Ans.   625  cents. 

NOTE.  To  change  dollars  and  cents  to  cents,  or  dollars,  cents,  and 
mills  to  mills,  remove  the  decimal  point  and  the  sign,  $. 

5.  Change  $63.045  to  mills.         Ans.   63045  mills. 

6.  Change  16  cents  to  mills. 

7.  Reduce  $3.008  to  mills. 

8.  In  89  cents  how  many  mills  ? 

162*     Conversely, 

To  change  cents  to  dollars,  divide  by  100  ;  that  is,  point  off 
two  figures  from  the  right. 

To  change  mills  to  dollars,  point  off  three  figures. 
To  change  mills  to  cents,  point  off  one  figure. 

How  are  dollars  changed  to  cents  ?  to  mills  ?  How  are  cents  changed 
to  mills?  How  are  cents  changed  to  dollars  ?  Mills  to  dollars  ?  to  cents? 


142  DECIMAL  CURRENCY. 

EXAMPLES    FOR    PRACTICE. 

1.  Change  875  cents  to  dollars.  Ans.    $8.75. 

2.  Change  1504  cents  to  dollars. 

3.  In  13875  cents  how  many  dollars  ? 

4.  In  16525  mills  how  many  dollars? 

5.  Reduce  524  mills  to  cents. 

6.  Reduce  6524  mills  to  dollars. 

ADDITION. 

1G3.  1.  A  man  bought  a  cow  for  21  dollars  50  cents,  a 
horse  for  125  dollars  37^  cents,  a  harness  for  46  dollars  75  cents, 
and  a  carriage  for  210  dollars ;  how  much  did  he  pay  for  all  ? 

OPERATION. 

$     21  50  ANALYSIS.     Writing  dollars  under  dol- 

125*375  ^ais'  cents  un(ler  cents,  &c.,  so  that  the 
decimal  points  shall  stand  under  each 
other,  we  add  and  point  off  as  in  addition 
of  decimals.  Hence  the  following 

Ans.   $403.625 

RULE.     I.    Write  dollars  under  dollars,  cents  under  cents,  fyc. 
II.  Add  as  in  simple  numbers,  and  place  the  point  in  the 
amount  as  in  addition  of  decimals. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  sum  of  50  dollars  7  cents,  1000  dollars  75 
cents,  60  dollars  3  mills,  18  cents  4  mills,  1  dollar  1  cent,  and 
25  dollars  45  cents  8  mills  ?  Ans.   $1137.475. 

3.  Add  364  dollars  54  cents  1  mill,  486  dollars  6  cents,  93 
dollars   9   mills,  1742  dollars  80  cents,  3  dollars  27  cents  6 
mills.  Ans.    $2689.686. 

4.  Add  92  cents,  10  cents  4  mills,  35  cents  7  mills,  18  cents 
6  mills,  44  cents  4  mills,  12£  cents,  and  99  cents.  Ans.  $3.126. 


Explain  the  process  of  addition  of  decimal  cxirrency .    Rule,  first  step. 
Second. 


SUBTRACTION.  143 

5.  A  farmer  receives  89  dollars  74  cents  for  wheat,  13  dol- 
lars 3  cents  for  corn,  6  dollars  374-  cents  for  potatoes,  and  19 
dollars    62^   cents   for  oats;    what  does  he  receive   for  the 
whole?  Am.   $128.77. 

6.  A  lady  bought  a  dress  for  9  dollars  17  cents,  trimmings 
for  87^  cents,  a  paper  of  pins  for  6^  cents,  some  tape  for  4 
cents,  some  thread  for  8  cents,  and  a  comb  for  1 1  cents ;  what 
did  she  pay  for  all  ?  Am.    $10.3375. 

7.  Paid  for  building  a  house  $2175.75,  for  painting  the 
same  $240.37=1,  for  furniture  $605.40,  for  carpets  $140.12£; 
what  was  the  cost  of  the  house  and  furnishing  ? 

8.  Bought  a  ton  of  coal  for  $6.08,  a  barrel  of  sugar  for 
$26.625,  a  box  of  tea  for  $16,  and  a  barrel  of  flour  for  $7.40  ; 
what  was  the  cost  of  all  ? 

9.  A  merchant  bought  goods  to  the  amount  of  $7425.50 ; 
he   paid   for   duties   on   the   same   $253.96,  and  for  freight 
$170.09  ;  what  was  the  entire  cost  of  the  goods  ? 

10.  I  bought  a  hat  for  $3.62±,  a  pair  of  shoes  for  $1|,  an 
umbrella  for  $lf ,  a  pair  of  gloves  for  $.62|,  and  a  cane  for 
$.87£ ;  what  was  the  cost  of  all  my  purchases  ?     Am.   $8.25. 

SUBTRACTION. 

164.    1.   A  man,  having  $327.50,  paid  out    $186.75  for 
a  horse ;  how  much  had  he  left  ? 

OPERATION.  ANALYSIS.     Writing  the  less  number  un- 

$327.50          der  the  greater,  dollars  under  dollars,  cents 

186  75          under  cents,  &c.,  we  subtract  and  point  off 

in  the  result  as  in  subtraction  of  decimals. 

Am.    $140.75          Hence  the  following 

RULE.     I.    Write  the  subtrahend  under  the  minuend,  dollars 
under  dollars,  cents  under  cents,  fyc. 

11.  Subtract  as  in  simple  numbers,  and  place  the  point  in 
the  remainder,  as  in  subtraction  of  decimals. 


Explain  the  process  of  subtraction.      Give  rule,  first  step.     Second. 


144  DECIMAL   CURRENCY. 

EXAMPLES    FOR    PRACTICE. 

2.  From  $365  dollars  5   mills  take  267  dollars  1  cent  8 
mills.  Ans.    $97.987. 

3.  From  50  dollars  take  50  cents.  Ans.    $49.50. 

4.  From  100  dollars  take  1  mill.  Ans.    $99.999. 

5.  From  1000  dollars  take  3  cents  7  mills. 

6.  A  man  bought  a  farm  for  $1575.24,  and  sold  it  for 
$1834.16;  what  did  he  gain  ?  Ans.    $258.92. 

7.  Sold  a  horse  for  145  dollars  27  cents,  which  is  37  dol- 
lars 69  cents  more  than  he  cost  me ;  what  did  he  cost  me  ? 

8.  A  merchant  bought  flour  for  $5.62 J  a  barrel, and  sold 
it  for  $6.84  a  barrel ;  how  much  did  he  gain  on  a  barrel  ? 

9.  A  gentleman,  having  $14725,  gave  $3560  for  a  store, 
and  $7015.87^-  for  goods ;  how  much  money  had  he  left  ? 

10.  A  lady  bought  a  silk  dress  for  $13J,  a  bonnet  for  $5^,  a 
pair  of  gaiters  for  $lf ,  and  a  fan  for  $| ;  she  paid  to  the  shop- 
keeper a  twenty  dollar  bill  and  a  five  dollar  bill ;  how  much 
change  should  he  return  to  her?  Ans.    $3.75. 

NOTE.   Reduce  the  fractions  of  a  dollar  to  cents  and  mills. 

11.  A  gentleman  bought  a  pair  of  horses  for  $480,  a  har- 
ness for  $80.50,  and  a  carriage  for  $200  less  than  he  paid  for 
both  horses  and  harness ;  what  was  the  cost  of  the  carriage? 

Ans.    $360.50. 

MULTIPLICATION. 

165.    1.   If  a  barrel  of  flour  cost  $6.375,  what  will  85 
barrels  cost  ? 

OPERATION. 

$6.375  ANALYSIS.    We  multiply  as  in   simple 

85          numbers,  always  regarding  the  multiplier 
as  an  abstract  number,  and  point  off'  from 
the  right  hand  of  the  result,  as  in  multipli- 
51000  cation  of  decimals.     Hence  the  following 

Ans.   $541.875 


Give  analysis  for  multiplication  in  decimal  currency. 


DIVISION.  145 

RULE.  Multiply  as  in  simple  numbers,  and  place  the  point 
in  the  product,  as  in  multiplication  of  decimals. 

EXAMPLES    FOR    PRACTICE. 

2.  If  a  cord  of  wood  be  worth  $4.275,  what  will  300  cords 
be  worth?  4ns.  $1282.50. 

3  What  will  175  barrels  of  apples  cost,  at  $2.45  per  bar- 
re!  ?  An*.  $428.75. 

4.  What  will  800  barrels  of  salt  cost,  at  $1.28  per  barrel? 

5.  A  grocer  bought  372  pounds  of  cheese  at  $.15  a  pound, 
434  pounds  of  coffee  at  $.12|  a  pound,  and  16  bushels  of  pota- 
toes at  $.33  a  bushel ;  what  did  the  whole  cost  ? 

6.  A  boy,  being  sent  to  purchase  groceries,  bought  3  pounds 
of  tea  at  56  cents  a  pound,  15  pounds  of  rice  at  7  cents  a 
pound,  27  pounds  of  sugar  at  8  cents  a  pound ;   he  gave  the 
grocer  5  dollars ;  how  much  change  ought  he  to  receive  ? 

7.  A  farmer  sold  125  bushels  of  oats  at  $.37^-  a  bushel, 
and  received  in  payment  75  pounds  of  sugar  at  $.09  a  pound, 
12  pounds  of  tea  at  $.60  a  pound,  and  the  remainder  in  cash; 
how  much  cash  did  he  receive  ?  Ans.    $32.92£. 

8.  A  man*  bought  150  acres  of  land  for  $3975  ;  he  after- 
ward sold  80  acres  of  it  at  $32.50  an  acre,  and  the  remainder 
at  $34.25  an  acre ;  how  much  did  he  gain  by  the  transaction  ? 

Ans.  $1022.50. 

i 
DIVISION. 

166.    1.   If  125  barrels  of  flour  cost  $850,  how  much 
will  1  barrel  cost  ? 

OPERATION.  ANALYSIS.    We  divide  as  in 

125  )  $850.00  (  $6.80,  Ans.       simple  numbers,  and  as  there 

is  a  remainder  after  dividing 
the  dollars,  we  reduce  the  div- 
idend to  cents,  by  annexing  two 
ciphers,  and  continue  the  di- 
vision. Hence  the  following 

Rule.      Give  rule  for  division  in  decimal  currency. 
M 


146  DECIMAL   CURRENCY. 

RULE.  Divide  as  in  simple  numbers,  and  place  the  point  in 
the  quotient,  as  in  division  of  decimals. 

NOTES.  1.  In  business  transactions  it  is  never  necessary  to  carry 
the  division  further  than  to  mills  in  the  quotient. 

2.  If  the  dividend  will  not  contain  the  divisor  an  exact  number  of 
tbnes,  ciphers  may  be  annexed,  and  the  division  continued  as  in  divis- 
ion of  decimals.  In  this  case  it  is  always  safe  to  reduce  the  dividend 
to  mills,  or  to  3  more  decimal  places  than  the  divisor  contains,  be- 
fore commencing  the  division. 

EXAMPLES    FOR   PRACTICE. 

2.  If  33  gallons  of  oil  cost  $41.25,  what  is  the  cost  per  gal- 
lon ?  Am.   $1.25. 

3.  If  27  yards  of  broadcloth  cost  $94.50,  what  will  1  yard 
cost? 

4.  If  64  gallons  of  wine  cost  $136j  what  will  1  gallon  cost? 

Ans.   $2.125. 

5.  At  12  cents  apiece,  how  many  pine-apples  can  be  bought 
for  $1.32?  Ans.  11. 

6.  If  1  pound  of  tea  cost  54  cents,  how  many  pounds  can 
be  bought  for  $405  ? 

7.  If  a  man  earn  $180  in  a  year,  how  mucji  does  he  earn 
a  month  ? 

8.  If  100  acres  of  land  cost  $2847.50,  what  will  1  acre 
cost?  Ans.'  $28.475. 

9.  What  cost  1  pound  of  beef,  if  894  pounds  cost  $80.46? 

Ans.   $.09. 

10.  A  farmer  sells  120  bushels  of  wheat  at  $1.12£  a  bushel, 
for  which  he  receives  27  barrels  of  flour ;  what  does  the  flour 
cost  him  a  barrel  ? 

11.  A  man  bought  4  yards  of  cloth  at  $3.20  a  yard,  and 
37  pounds  of  sugar  at  $.08  a  pound ;  he  paid  $6.80  in  cash, 
and  the  remainder  in  butter  at  $.16  a  pound ;  how  many  pounds 
of  butter  did  it  take  ?  Ans.  56  pounds. 

12.  A  man  bought  an  equal  number  of  calves  and  sheep, 
paying  $166.75  for  them ;  for  the  calves  he  paid  $4.50  a  head, 
and  for  the  sheep  $2.75  a  head ;  how  many  did  he  buy  of  each 
kind?  Ans.   23. 


OF 


APPLICATIONS. 


13.  If  154  pounds  of  sugar  cost  $18.48,  W 
cost? 

14.  A  merchant  bought  14  boxes  of  tea  for  $560;  it  being 
damaged  he  was  obliged  to  lose  $106.75  on  the  cost  of  it  ; 
how  much  did  he  receive  a  box  ?  Ans.    $32.37£. 

ADDITIONAL  APPLICATIONS. 

CASE    I. 

167.  To  find*  the  cost  of  any  number  or  quantity, 
when  the  price  of  a  unit  is  an  aliquot  part  of  one  dollar. 

168.  An  Aliquot  Part  of  a  number  is  such  a  part  as  will 
exactly  divide  that  number;  thus,  2,  3,  and  7^  are  aliquot 
parts  of  15. 

NOTE.    An  aliquot  part  may  be  a  whole  or  a  mixed  number,  while  a 
factor  must  be  a  whole  number. 


ALIQUOT    PARTS    OF    ONE    DOLLAR. 


50  cents  = 
33  £  cents  = 
25  cents  = 
20  cents  = 
16f  cents  = 


of  1  dollar, 

of  1  dollar, 

of  1  dollar, 

of  1  dollar, 

of  1  dollar. 


121  cents  =  £  of  1  dollar. 
10    cents  —  fa  of  1  dollar. 

8£  cents  =  fa  of  1  dollar. 

6£  cents  =  -fa  of  1  dollar. 

5    cents  =  ^  of  1  dollar. 


1.  What  will  be  the  cost  of  3784  yards  of  flannel,  at  25 
cents  a  yard  ? 

OPERATION.          ANALYSIS.     If  the  price  were   $1   a  yard, 

4  ^  3784         ^e  cos^  wou^  be  as  many  dollars  as  there  are 

yards.     But  since  the  price  is  \  of  a  dollar  a 

yard,  the  whole  cost  will  be  \  as  many  dollars 

as  there  are  yards  ;  or,  }  of  3784  =  3784  -^  4  =  $946.     Hence  the 

RULE.     Take  such  a  fractional  part  of  the  given  number  as 
the  price  is  part  of  one  dollar. 

EXAMPLES    FOR    PRACTICE. 

2.  What  cost  963  bushels  of  oats,  at  33£  cents  per  bushel? 

Ans.    $321. 

Case  I  is  what  ?     What  is  an  aliquot  part  of  a  dollar  ?     Give  ex- 
planation.   Rule. 


148  DECIMAL   CURRENCY. 

3.  What  cost  478  yards  of  delaine,  at  50  cents  per  yard  ? 

4.  What  cost  4266  yards  of  sheeting,  at  8£  cents  a  yard? 

Ans.    $355.50. 

5.  What   cost  1250  bushels   of  apples,  at  12  i   cents  per 
feushel?  Ans.    $156.25. 

6.  What  cost  3126  spools  of  thread,  at  6|  cents  per  spool? 

Ans.    $195.375. 

7.  At  16§  cents  per  dozen,  what  cost  1935  dozen  of  eggs? 

.     Ans.   322.50. 

8.  What  cost  56480  yards  of  calico,  at  12£  per  yard  ? 

9.  At  20  cents    each  what  will   be   the  cost  of  1275  salt 
barrels?  Ans.   $255. 

CASE    II. 

169.  The  price  of  one  and  the  quantity  being  given, 
to  find  the  cost. 

1.  How  much  will  9  barrels  of  flour  cost,  at  $6.25  per 
barrel  ? 

OPERATION.  ANALYSIS.     Since  one  barrel  cost  $6.25,9 

$6.25          barrels  will  cost  9  times  $6.25,  and  $6.25  X 
9  =  $56.25.     Hence 


Ans.    $56.25 

RULE.     Multiply  the  price  of  one  by  the  quantity. 

EXAMPLES    FOR   PRACTICE. 

2.  If  a  pound  of  beef  cost  9  cents,  what  will  864  pounds 
cost?  Ans.   $77.76. 

3.  What  cost  87  acres  of  government  land,  at  $1.25  per 
acre? 

4.  What  cost  400  barrels  of  salt,  at  $1.45  per  barrel  ? 

Ans,    $580. 

5    What  cost  16  chests  of  tea,  each  chest  containing  52 
pounds,  at  44  cents  per  pound  ? 

Case  n  is  what  ?     Give  explanation.     Rule. 


APPLICATIONS.  149 

CASE  III. 

170 .  The  cost  and  the  quantity  being  given,  to  find 
the  price  of  one. 

1.  If  30  bushels  of  corn  cost  $20.70,  what  will  1  bushel 
cost? 

OPERATION.  ANALYSIS.     If  30  bushels  cost  $20.70,   1 

310  )  $210  70        bushel  will  cost  -fa  of  $20.70;  and  $20.70  -f- 

30  =  $.69.     Hence, 
$.69 

RULE.     Divide  the  cost  by  the  quantity. 

EXAMPLES    FOR    PRACTICE. 

2.  If  25  acres  of  land  cost  $175,  what  will  1  acre  cost? 

3.  If  48  yards  of  broadcloth  cost  $200,  what  will  1  yard 
cost?  Am.   $4.1 6§. 

4.  If  96  tons  of  hay  cost  $1200,  what  will  1  ton  cost? 

5.  If  10  Unabridged  Dictionaries  cost  $56.25,  what  will  1 
cost?  Ans.    $5.62£. 

6.  Bought  18  pounds  of  tea  for  $11.70 ;  what  was  the  price 
per  pound  ?  Ans.    $.65. 

7.  If  53  pounds  of  butter  cost  $10.07,  what  will  1  pound 
cost? 

8.  A  merchant  bought  800  barrels  of  salt  for  $1016 ;  what 
did  it  cost  him  per  barrel  ? 

9.  If  343  sheep  cost  $874.65,  what  will  1  sheep  cost  ? 

•  Ans.   $2.55. 

10.  If  board  for  a  family  be  $684.37£  for  1  year,  how  much 
is  it  per  day?  Ans.   $1.87£. 

CASE   IV. 

171.  The  price  of  one  and  the  cost  of  a  quantity 
being  given,  to  find  the  quantity. 

1.   At  $6  a  barrel  for  flour,  how  many  barrels  can  be  bought 

for  $840  ? 

Case  III  is  what  ?     Give  explanation.    Rule.     Case  IY  is  what  ? 


150  DECIMAL   CURRENCY. 

OPERATION.  ANALYSIS.     Since  $6  will  buy  1  barrel 

6  )  840  of  flour>  $840  will  buy  £  as  many  barrels 

.      ~T7I  as  there  are  dollars,  or  as  many  barrels  as 

Am.    ]  reis.         $6  is  contained  times  in  $840;  840-^-6 

z=  140  barrels.     Hence, 

RULE.     Divide  the  cost  of  the  quantity  by  the  price  of  one. 

EXAMPLES    FOR    PRACTICE. 

2.  How  many  dozen  of  eggs  can  be  bought  for  $5.55,  if  one 
dozen  cost  $.15  ?  Am.   37  dozen. 

3.  At  $12  a  ton,  how  many  tons  of  hay  can  be  bought  for 
$216?  Ans.   18  tons. 

4.  How  many  bushels  of  wheat  can  be  bought  for  $2178.75, 
if  1  bushel  cost  $1.25  ?  Ans.   1743  bushels. 

5.  A  dairyman  expends  $643.50  in  buying  cows  at  $19£ 
apiece  ;  how  many  cows  does  he  buy  ?  Ans.   33  cows. 

6.  At  $.45  per  gallon,  how  many  gallons  of  molasses  can 
be  bought  for  $52.65  ? 

7.  A  drover  bought  horses  at  $264  a  pair;  how  many 
horses  did  he  buy  for  $6336  ? 

8.  At  $65  a  ton,  how  many  tons  of  railroad  iron  can  be 
bought  for  $117715?  Ans.   1811  tons. 

CASE   V. 

172.    To  find  the  cost  of  articles  sold  by  the  100, 
1000,  &c. 

1.   What  cost  475  feet  of  timber,  at  $5.24  per  100  feet  ? 

FIRST  OPERATION. 

$5  24  ANALYSIS.    If  the  price  were  $5.24  per 

475         foot>  the  cost  °f  475  feet  would  be  475  X 

$5.24  — $2489.      But   since    $5.24  is   the 

2620         price  of  100  feet,  $2489  is  100  times  the  true 

3668  value.     Therefore,  to  obtain  the  true  value, 

2096  we  divide  $2489  by  100,  which  we  may  do 

i ™  \  ^QOAA         by  cuttin£  off  two  %ures  from  the  riSht»  and 
100  )  $2489.00         the  result  is  $24.89.     Or, 

Ans.   $24.89 
Give  explanation.    Rule.    Case  V  is  what  ?    Give  first  explanation. 


APPLICATIONS.  151 

SECOND  OPERATION.      ANALYSIS.     Since  1  foot  costs  T^,  or.  01, 
$5.24  of>  $5.24,  475  feet  will  cost  f£|,  or  4.75  times 

4  75  $5.24,  which  is  $24.89. 

NOTE.     For  the  same  reasons,  when  the  price 


ofi,  is  per  thmisand,  we  divide  the  product  by  1000, 

or,  which  is  more  convenient  in  practice,  we  re- 

2096  duce  the  given  quantity  to  thousands  and  deci- 

mals of  a  thousand,  by  pointing  off  three  figures 

$24.8900  from  the  right  hand.     Hence  the 

RULE.  I.  Reduce  the  given  quantity  to  hundreds  and  deci- 
mals of  a  hundred,  or  to  thousands  and  decimals  of  a  thousand. 

II.  Multiply  the  price  by  the  quantity,  and  point  off  in  the 
result  as  in  multiplication  of  decimals. 

NOTE.  The  letter  C  is  used  to  indicate  hundreds,  and  Mto  indicate 
thousands. 

EXAMPLES    FOR   PRACTICE. 

2.  What  will  42650  bricks  cost,  at  $4.50  per  M  ? 

Ans.   $191.925. 

3.  What  is  the  freight  on  2489  pounds  from  Boston  to  New 
York,  at  $.85  per  100  pounds  ?  Ans.    $21.156+. 

4.  What  will  7842   feet  of   pine  boards  cost,  at    $17.25 
perM?  Ans.    $135.274+. 

5.  What  cost  2348  pine-apples,  at  $12£  per  100  ? 

6.  A  broom  maker  bought  1728  broom-handles,  at  $3  per 
1000  ;  how  much  did  they  cost  him  ? 

7.  What  is  the  cost  of  2400  feet  of  boards,  at  $7  perM; 
865  feet  of  scantling,  at  $5.40  per  M;  and  1256  feet  of  lath,  at 
$.80  per  C?  Ans.   $31.519. 

8.  What  will  be  t£e  cost  of  1476  pounds  of  beef,  at  $4.374- 
per  hundred  pounds  ? 

CASE    VI. 

173.  -To  find  the  cost  of  articles  sold  by  the  ton  of 
2000  pounds. 

1  .  How  much  will  2376  pounds  of  hay  cost,  at  $9.50  per  ton  ? 
Give  second  explanation.  Rule,  first  step.  Second.  Case  VI  is  what  ? 


152  DECIMAL   CURRENCY. 

OPERATION.  ANALYSIS.     Since  1  ton,  or  2000  pounds,  cost 

2  )  $9.50  $9.50, 1000  pounds,  or  -|  ton,  will  cost  ^  of  $9.50, 

^ —  or   $9.50  -f-  2  =  $4,75.     One  pound  will  cost 

TJrg-,  or  .001,  of  $4.75,  and  2376  pounds  will 

2.376  cost  f £££,  or  2.376  times  $4.75,  which  is  $1 1.286. 

$11.28600       Hence> 

RULE.  I.  Divide  the  price  of  1  ton  by  2,  and  the  quotient 
will  be  the  price  of  1000  pounds. 

II.  Multiply  this  quotient  by  the  given  number  of  pounds 
expressed  as  thousandths,  as  in  Case  V. 

EXAMPLES    FOR   PRACTICE. 

2.  At  $7  a  ton,  what  will  1495  pounds  of  hay  cost  ? 

Ans.   $5.2325. 

3.  At  $8.75  a  ton,  what  cost  325  pounds  of  hay  ? 

Ans.   $1.421+. 

4.  What  is  the  cost  of  3142  pounds  of  plaster,  at  $3.84  per 
ton?  Ans.   $6.032+. 

5.  What  is  the  cost  of  1848  pounds  of  coal,  at  $5.60  per 
ton? 

6.  Bought  125  sacks  of  guano,  each  sack  containing  148 
pounds,  at  $18  a  ton;  what  was  the  cost? 

7.  What  must  be  paid  for  transporting  31640  pounds  of 
railroad  iron  from  Philadelphia  to  Richmond,  at  $3.05  per 
ton?  Ans.    $48.251. 

BILLS. 

17" 4:.  A  Bill,  in  business  transactions,  is  a  written  state- 
ment of  articles  bought  or  sold,  together  with  the  prices  of 
each,  and  the  whole  cost. 

Find  the  cost  of  the  several  articles,  and  the  amount  or 
footing  of  the  following  bills.  , 


Give  explanation.     Rule.     What  is  a  bill  ?     Explain  the  manner  of 
making  out  a  bill. 


BILLS.  153 


Mr.  JOHK  RICE,  NEW  YoKK'  June  20>  1859' 

Bo't.  of  BALDWIN  &  SHERWOOD, 

7  yds.  Broadcloth,  fa)  $3.60 

9    "     Satinet,  "     1.12£ 

12    «     Vesting,  «       .90 

24    "     Cassimere,  "     1.37£ 

32    "     Flannel,  «       .65  _ 

„    ,  ,  D  $99.925 

See  d  Payment,          BALDWIN  &  SHERWOOD. 

(2-) 

DANIEL  CHAPMAN  &  Co.,  BosTON>  Jan'  ^  1860' 

^o'^.  O/PALMEK  &  BROTHER. 

67  pairs  Calf  Boots,  (a)  $3.75 
108  "  Thick  "  "  2.62 
75  "  Gaiters,  "  1.12 
27  "  Buskins,  "  .86 
35  «  Slippers,  "  .70 
50  «  Rubbers,  «  1.04  _ 

$717.93 
Jtec  a.  Jraymentj 

PALMER  &  BROTHER, 

By  GEO.  BAKER. 

(3.) 

G.  B.  GEANNIS,  CHARLESTON,  Sept.  6,  1859. 

Bo't.  of  STEWART  &  HAMMOND, 
325  Ibs.  A.  Sugar,       (a)     $.07 
148    "   B.      «  "         .06^ 

286    «   Rice,  "         .05 

95    «   O.  J.  Coffee,   "        .12^- 
50  boxes  Oranges,     "       2.75 
75      "      Lemons,      "       3.62^ 
12      "      Raisins,       "       2.85  ___ 

501.75 
Payment,  by  note  at  4  mo. 

STEWART  &  HAMMOND. 


154  DECIMAL  CURRENCY. 


Messrs.  OSBORN  &  EATON,      ST"  LouIS>  Oct'  15>  1858' 
Bo't.  of  ROB'T.  H.  CARTER  &  Co., 
20000  feet  Pine  Boards  fa>   $15     per  M. 


7500    « 

Plank, 

t( 

9.50     « 

10750    « 

Scantling, 

u 

6.25     « 

3960    « 

Timber, 

(i 

2.62£  « 

5287    « 

a 

a 

3.00     " 

$464.6935 
Hec'd.  Payment, 

KOB'T.  H.  CARTER  &  Co. 

(5.) 

Mr.  J.  C.  SMITH,  CINCINNATI,  May  3,  1861. 

Bo't.  of  SILAS  JOHNSON, 

25  Ibs.      Coffee  Sugar, 

5  "        Y.  H.  Tea, 

26  "        Mackerel, 
4  gal.      Molasses, 

46  yds.     Sheeting, 

30    «        Bleached  Shirting, 

6  skeins  Sewing  Silk,  ' 
4  doz.     Buttons, 

Chgd.  in  a|c. 


PROMISCUOUS    EXAMPLES. 

1.  What  will  62.75  tons  of  potash  cost,  at  $124.3^  per  ton  ? 

~Ans.   $7802.9625. 

2.  "What  cost  15  pounds  of  butter,  at  $.17  a  pound? 

Aw.  $2.55. 

3.  A  cargo  of  corn,  containing  2250  bushels,  was  sold  for 
$1406.25  ;  what  did  it  sell  for  per  bushel  ?  Aw.    $£ . 


PROMISCUOUS  EXAMPLES.  155 

4.  If  12  yards  of  cloth  cost  $48.96,  what  will  one  yard 
cost  ? 

5.  A  traveled  325  miles  by  railroad,  and  C  traveled  .45  of 
that  distance  ;  how  far  did  C  travel  ?      Ans.    146.25  miles. 

6.  If  36.5   bushels  of  corn  grow  on  one  acre,  how  ma^iy 
acres  will  produce  657  bushels  ?  Ans.    18  acres. 

7.  Bought  a  horse  for  $105,  a  yoke  of  oxen  for  $125,  4 
cows  at  $35  apiece,  and  sold  them  all  for  $400 ;  how  much 
was  gained  or  lost  in  the  transaction  ? 

8.  A  man  bought  28  tons  of  hay  at  $19  a  ton,  and  sold  it 
at  $15  a  ton  ;  how  much  did  he  lose  ?  Ans.    $112. 

9.  If  a  man  travel  4f  miles  an  hour,  in  how  many  hours 
can  he  travel  34£  miles  ?  Ans.   7.5  hours. 

10.  At  $.31£  per  bushel,  how  many  bushels  of  potatoes 
can  be  bought  for  $9  ?  Ans.    28.8  bushels. 

11.  If  a  man's  income  be  $2000  a  year,  and  his  expenses 
$3.50  a  day,  what  will  he  save  at  the  end  of  a  year,  or  365 
days? 

1 2.  A  merchant  deposits  in  a  bank,  at  one  time,  $687.25, 
and  at  another,  $943.64  ;  if  he  draw  out  $875.29,  how  much 
will  remain  in  the  bank  ? 

13.  Bought  288  barrels  of  flour  for  $1728,  and  sold  one 
half  the  quantity  for  the  same  price  I  gave  for  it,  and  the  other 
half  for  $8  per  barrel ;  how  much  did  I  receive  for  the  whole  ? 

Ans.    $2016. 

14.  What  will  eight  hundred  seventy-five  thousandths  of  a 
cord  of  wood  cost,  at  $3.75  per  cord  ?          Ans.    $3.281+. 

15.  A  drover  bought  cattle  at  $46.56  per  head,  and  sold 
them  at  $65.42  per  head,  and  thereby  gained  $3526.82 ;  how 
many  cattle  did  he  buy  ?  Ans.    187. 

16.  If  36.48  yards  of  cloth  cost  $54.72,  what  will  14.25 
yards  cost?  Ans.   $21.375. 

17.  A  house  cost  $3548,  which  is  4  times  as  much  as  the 
furniture  cost ;  what  did  the  furniture  cost  ?        Ans.    $887. 

18.  How  many  bushels  of  onions  at  $.82  per  bushel,  can 
be  bought  for  $112.34? 


156  DECIMAL   CURRENCY. 

19.  If  46  tons  of  iron  cost  $3461.50,  what  will  5  tons  cost? 

20.  A  gentleman  left  his  widow  one  third  of  his  property, 
worth  $24000,  and  the  remainder  was  to  be  divided  equally 
among  5  children  ;  how  much  was  the  portion  of  each  child  ? 

Ans.    $3200. 

21.  A  man  purchased  one  lot,  containing  1 60  acres  of  land,  at 
$1.25  per  acre ;  and  another  lot,  containing  80  acres,  at  $5  per 
acre  ;  he  sold  them  both  at  $2.50  per  acre ;  what  did  he  gain 
or  lose  in  the  transaction  ? 

22.  A  druggist  bought  54  gallons  of  oil  for  $72.90,  and 
lost  6  gallons  of  it  by  leakage.     He  sold  the  remainder  at 
$1.70  per  gallon;  how  much  did  he  gain  ?         Ans.    $8.70. 

23.  A  miller  bought  122J-  bushels  of  wheat  of  one  man, 
and  75£  bushels  of  another,  at  $.93J  per  bushel.     He  sold  60 
bushels  at   a  profit  of  $12.50  ;  if  he  sell  the  remainder  at 
$.81£  per  bushel,  what  will  be  his  entire  gain  or  loss  ? 

Ans.    $4.718+  loss. 

24.  A  laborer  receives  $1.40  per  day,  and  spends  $.75  for 
his  support ;  how  much  does  he  save  in  a  week  ? 

25.  How  many  pounds  of  butter,  at  $.16  per  pound,  must 
be  given  for  39  yards  of  sheeting,  at  $.08  a  yard? 

Ans.    19£  pounds. 

26.  What  cost  23487  feet  of  hemlock  boards,  at  $4.50  per 
1000  feet?  Ans.   $105.6915. 

27.  A  man  has  an  income  of  $1200  a  year;  how  much 
must  he  spend  per  day  to  use  it  all  ? 

28.  Bought  28  firkins  of  butter,  each  containing  5  6  pounds, 
at  $.17  per  pound  ;  what  was  the  whole  cost  ? 

29.  A  merchant  bought  1 6  bales  of  cotton  cloth,  each  bale 
containing  13  pieces,  and  each  piece  26  yards,  at  $.07  per 
yard ;  what  did  the  whole  cost  ?  Ans.    $378.56. 

30.  What  cost  4868  bricks, at  $4.75  per  M  ? 

31.  A  farmer  sold  27  bushels  of  potatoes,  at  $.33£  per 
bushel ;  28  bushels  of  oats,  at  $.25  per  bushel ;  and  19  bush- 
els of  corn,  at  $.50  per  bushel ;  what  did  he  receive  for  the 
whole?  Ans.   $25.50. 


PROMISCUOUS    EXAMPLES.  157 

32.  John    runs   32    rods   in  a  minute,  and  Henry  pursues 
him  at  the  rate  of  44  rods  in  a  minute;  how  long  will  it  take 
Henry  to  overtake  John,  if  John  have  8  minutes  the  start  ? 

Am.    21^  minutes. 

33.  If  4£  barrels  of  flour  cost  $32.3,  what  will  7±  barrels 
cost?      •  Am.    $51. 

34.  If  .875  of  a  ton  of  coal  cost  $5.635,  what  will  9£  tons 
cost?  Am.    $59.57. 

35.  For  the  first  three  years  of  business,  a  trader  gained 
$1200.25  a  year;  for  the  next  three,  he  gained  $1800.62  a 
year,  and  f#r  the  next  two  he  lost  $950.87  a  year ;  supposing 
his  capital  at  the  beginning  of  trade  to  have  been  $5000,  what 
was  he  worth  at  the  end  of  the  eighth  year  ?  Am.  $12100.82. 

36.  What  will  be  the  cost  of  18640  feet  of  timber,  at  $4.50 
per  100  ?  Ans.    $838.80. 

2i 

37.  Reduce  57  to  a  decimal  fraction.  Ans.    .78125. 

6% 

38.  What  will  1375  pounds  of  potash  cost,  at  $96.40  per 
ton?  Ans.    $66.275. 

39.  Reduce  .5625  to  a  common  fraction.  Ans.   T9^. 

40.  Reduce  ^-,  .62^,  .37^,  f ,  to  decimals,  and  find  their 
sum.  Ans.    1.464375. 

41.  A  man's  account  at  a  store  stands  thus : 

Dr.  Cr. 

$4.745  $2.76£ 

2.62£  1.245 

1.27  .62£ 

*          .45  3.45 

5.28£  1.87J 

What  is  due  the  merchant  ?  Ans.    $4.41£. 

42.  A  gardener  sold,  from  his  garden,  120  bunches  of  on- 
ions at  $.12^-  a  bunch,  18  bushels  of  potatoes  at  $.62£  per 
bushel,  47  heads  of  cabbage  at  $.07  a  head,  6  dozen  cucum- 
bers at  $.18  a  dozen;  he  expended  $1.50  in  spading,  $1.27 
for  fertilizers,  $1.87  for  seeds,  $2.30  in  planting  and  hoeing; 
what  were  the  profits  of  his  garden  ?  Ans.    $23.68. 

N 


158  REDUCTION. 


REDUCTION. 

175*  A  Compound  Number  is  a  concrete  number  whose 
value  is  expressed  in  two  or  more  different  denominations. 
(110.) 

176.  Reduction  is  the  process  of  changing  a  number  from 
one  denomination  to  another  without  altering  its  value. 

Reduction  is  of  two  kinds,  Descending  and  Ascending. 

177.  Reduction  Descending  is  changing  a  number  of  one 
denomination  to  another  denomination  of  less  unit  value  ;  thus, 
$1  =  10  dimes  =  100  cents  =  1000  mills. 

178.  Reduction  Ascending  is  changing  a  number  of  one 
denomination  to  another  denomination  of  greater  unit  value  ; 
thus,  1000  mills  =  100  cents  =  10  dimes  =$1. 

179.  A  Scale  is  a  series  of  numbers,  descending  or  as- 
cending, used  in  operations  upon  compound  numbers. 

NOTE.  In  simple  numbers  and  decimals,  the  scale  is  uniformly  10  ; 
in  compound  numbers  the  scales  are  varying. 

CURRENCY. 

180.  I.     UNITED  STATES  MONEY. 

TABLE. 

10  mills  (m.)  make  1  cent, ct. 

10  cents  "  1  dime, d. 

10  dimes  "  1  dollar, $. 

10  dollars  "  1  eagle, E. 

UNIT   EQUIVALENTS. 

ct.  m. 

,1.  1   ~  10 

$             1  =  10  =       100 

E.          1  =     10  =  100  =     1000 

1  =  10  =  100  =  1000  =  10000 
SCALE  —  uniformly  10. 

NOTE.  This  table  is  given  here  for  the  purpose  of  presenting  all  the 
compound  numbers  together ;  but  Decimal  Currency  has  been  previ- 
ously treated  of  so  fully,  that  additional  exercises  are  not  considered 
necessary. 

AVhat.  is  a  compound  number?  Reduction  i*  what?  Reduction 
descending  ?  Reduction  ascending  ?  A  scale  ?  Scale  of  simple  num- 
bers and  decimals  ?  Scale  of  compound  numbers  ?  Repeat  the  table 
of  U.  S.  money. 


COMPOUND   NUMBERS.  159 

II.     ENGLISH  MONEY. 
English  Money  is  the  currency  of  Great  Britain. 

TABLE. 

4  farthings  (far.  or  qr.)  make  1  penny, d. 

12  pence  "       1  shilling, s. 

20  shillings  "       1  pound  or  sovereign,.  .£,  or  sov. 

UNIT    EQUIVALENTS. 

d.  far. 

1—4 

£,orsov.       1   =      12   =      48 

1  —  20  =  240  =  960 
.SCALE  ascending,  4,  12,  20;  descending,  20,  12,  4. 

NOTES.  1.  The  equivalents  of  each  unit  in  all  lower  denominations 
may  easily  be  learned,  when  but  few  denominations  and  small  numbers 
are  used. 

2.  Farthings  are  generally  expressed  as  fractions  of  a  penny ;  thus, 
1  far.,  sometimes  called  1  quarter,  (qr.)  =£d. ;  3  far.  =  f  d. 

CASE   I. 

18S.    To  perform  reduction  descending. 
1.    Reduce  21  £  18  s.  10  d.  2  far.  to  farthings. 

OPERATION.  ANALYSIS.    We  multiply 

21  £  18s.  10  d.  2  far.         21  £  by  20,   because  there 
20  are  20  times  as  many  shil- 

lings as  pounds,  and  add 
18  shillings  to  the  product, 
making  438  shillings  in  21  £ 

5266  d.  ^ s<       e  next  muhiply  438 

M  shillings     by    12,    because 

there  are  12  times  as  many 

Ans.    21066  far.  pence  as  shillings,  and  add 

10  pence   to   the   product, 

making  5266  pence  in  21  £  18s.  10  d.  Lastly,  we  multiply  the  5266 
pence  by  4,  because  there  are  4  times  as  many  farthings  as  pence, 
and  add  2  farthings  to  the  product;  and  we  obtain  21066  farthings 
in  the  given  number.  Hence  the  following 

RULE.  I.  MULTIPLY  the  highest  denomination  of  the  given 
number  by  that  number  of  the  scale  which  ivill  reduce  it  to 
the  next  lower  denomination,  and  add  to  the  product  the  given 
number,  if  any,  of  that  lower  denomination. 

Repeat  the  table  of  English  money.  The  scale.  Case  I  is  what  r 
Give  explanation.  Rule. 


160  REDUCTION. 

II.  Proceed  in  the  same  manner  with  the  results  obtained  in 
each  lower  denomination,  until  the  reduction  is  brought  to  the 
denomination  required. 

CASE   II. 

183.   To  perform  reduction  ascending. 

I.  Reduce  21066  farthings  to  pounds. 

OPERATION.  ANALYSIS.    We  first  divide 

4)  21066  far.  the  21066  far.  by  4,  because 

there  are  \  as  many  pence  as 

12)5266d.  +  2far.  farthings,   and  we  find  that 

2|0  )  43|8  s.  +  10  d.  21066  far.  —  5266  d.  +  a  re- 

o7  f  _l_  i  Q  mainder  of  2  far.     We  next 

21  £.  •+-  18  s.  ^^  5266  d>  by  12>  because 

Ans.   21  £  18  s.  10  d.  2  far.         there  are  ^  as  many  shining8 

as  pence,  and  we  find  that  5266  d.  z=  438  s.  -j-  10  d.  Lastly,  we 
divide  the  438  s.  by  20,  because  there  are  -^  as  many  pounds  as 
shillings,  and  we  find  that  438  s.  =  21  £  + 18  s.  The  last  quotient 
with  the  several  remainders  annexed  in  the  order  of  the  succeeding 
denominations  gives  the  answer  21  £  18  s.  10  d.  2  far.  Hence  the 
following 

RULE.  I.  DIVIDE  the  given  number  by  that  number  of  the 
scale  which  will  reduce  it  to  the  next  higher  denomination. 

II.  Divide  the  quotient  by  the  next  higher  number  in  the 
scale ;  and  so  proceed  to  the  highest  denomination  required. 
The  last  quotient,  with  the  several  remainders  annexed  in  a 
reversed  order,  will  be  the  answer. 

NOTE.  Reduction  descending  and  reduction  ascending  mutually 
prove  each  other. 

EXAMPLES    FOR    PRACTICE. 

1.  In  14194  farthings  how  many  pounds? 

2.  In  14  £  15  s.  8  d.  2  far.  how  many  farthings  ? 

3.  Reduce  15  £  19  s.  11  d.  3  far.  to  farthings. 

4.  In  15359  farthings  how  many  pounds? 

5.  In  46  sov.  12s.  2  d.  how  many  pence  ? 

6.  In  11186  pence  how  many  sovereigns? 

Case  II  is  what  ?    Give  explanation.    Rule.    Proof,  how  performed  ? 


COMPOUND   NUMBERS. 


161 


WEIGHTS. 

184:.  Weight  is  a  measure  of  the  quantity  of  matter  a 
body  contains,  determined  according  to  some  fixed  standard. 
Three  scales  of  weight  are  used  in  the  United  States 
and  Great  Britain,  namely,  Troy,  Apothecaries',  and  Avoir- 
dupois. 

I.     TROY  WEIGHT. 

185.  Troy  Weight  is  used  in  weighing  gold,  silver,  and 
jewels ;  in  philosophical  experiments,  &c. 

TABLE. 

24  grains    (gr.)    make  1  pennyweight,,  .pwt.  or  dwt. 

20  pennyweights      "      1  ounce, oz. 

12  ounces  "      1  pound, Ib. 

UNIT   EQUIVALENTS. 

pwt.  gr. 

1  =       24 

ib.          1  —    20  =    480 
1  =  12  =  240  =  5760 
SCALE  —  ascending,  24,  20,  12;  descending,  12,  20,  24. 

EXAMPLES    FOR    PRACTICE. 


1.  How  many  grains  in  14  Ib. ' 
10  oz.  18  pwt.  22  gr.? 

OPERATION. 

14  Ib.  10  oz.  18  pwt.  22  gr. 
12 


178  oz. 
20 


3578  pwt. 
24 


14334 
7156 


2.    How    many   pounds    in 
85894  grains  ? 

OPERATION. 

24 )  85894  gr. 
20  )  3578  pwt.  +  22  gr. 
12  )178  oz.  +  15  pwt. 
14  Ib.  +  10oz. 

Ans.    141b.  lOoz.  18  pwt. 
22  gr. 


85894  gr.,  Ans. 

3.  In  5  Ib.  7  oz.  12  pwt.  9  gr.,  how  many  grains  ? 

4.  In  43457  grains  how  many  pounds  ? 

Define  weight.    Troy  weight.    Repeat  the  table.     Give  the  scale. 


162 


REDUCTION. 


5.  Reduce  41760  grains  to  pounds.         Ans.   7  Ib.  3  oz. 

6.  A  miner  had   14 Ib.   10  oz.  18  pwt.  of  gold  dust;  how 
much  was  it  worth  at  $.75  a  pwt.  ?  Ans.    $2683.50. 

7.  How  many  spoons,  each  weighing  2  oz.  15  pwt.,  can  be 
made  from  5  Ib.  6  oz.  of  silver  ?  Ans.    24. 

8.  A  goldsmith  manufactured  1  Ib.  1  pwt.  16  grs.  of  gold  into 
rings,  each  weighing  4  pwt.  20  gr. ;  he  sold  the  rings  for  $1.25 
apiece ;  how  much  did  he  receive  for  them  ?    Ans.    $62.50. 

II.     APOTHECARIES'  WEIGHT. 

186.  Apothecaries'  Weight  is  used  by  apothecaries  and 
physicians  in  compounding  medicines ;  but  medicines  are 
bought  and  sold  by  avoirdupois  weight. 

TABLE. 

20  grains  (gr.)  make  1  scruple, sc.  or  9- 

3  scruples  "      1  dram, dr.  or  3  • 


8  drams 
12  ounces 


1  ounce, oz.  or 

1  pound, Ib.  or 


UNIT    EQUIVALENTS. 

so.  gr. 

dr.  ^1    =        20 

oz  1  =      3  =      60 

lb.          1  —     8  =    24  —    480 
1  =  12  =  96  =  288  ==  5760 
SCALE  — ascending,  20,  3,  8,  12;  descending,  12,  8,  3,  20. 

EXAMPLES    FOR    PRACTICE. 


1.    How  many  gr.  in  12  ib 
85  35  19  15  gr.? 

OPERATION. 

12  ib  8§35  19  15  gr. 

12                   s 

2.    How  many  Ib  in  73175 
gr.? 

OPERATION. 

2|0)  7317)5  gr. 
3)36589  +  15gr. 

152  5 

8 

8  )  12193  +  19 
12  )  152  §  +  3  3 

1219  3 
3 

12tb  +  8§ 

3658  B 
20 

Ans.  12lb8§  3  3  19  15  gr. 

73175  gr.,  Ans. 

Define  apothecaries'  weight.    Repeat  the  table.     Give  the  scale. 


COMPOUND   NUMBERS.  163 

3.  In  1  Gib.  11  oz.  7  dr.  2  sc.  19  gr.,  how  many  grains? 

4.  Reduce  47  ft  6  §  4  5  to  scruples.        Am.    13692' sc. 

5.  How  many  pounds  of  medicine  would  a  physician  use  in 
one  year,  or  365  days,  if  he  averaged  daily  5   prescriptions 
of  20  grains  each  ?  Ans.    6  ib.  4  §  1  £). 

III.     AVOIRDUPOIS  WEIGHT. 

187.  Avoirdupois  Weight  is  used  for  all  the  ordinary  pur- 
poses of  weighing. 

TABLE. 

16  drams  (dr.)  make  1  ounce, oz. 

16  ounces  "       1  pound, Ib. 

100  Ib.                                "1  hundred  weight,  .  cwt. 
/     20  cwt.,  or  2000  Ibs.,      "       1  ton, T. 

UNIT    EQUIVALENTS. 

oz.  dr. 

Ib.  1   =  16 

cwt.  1  —        16  =        256 

T  1  =     100  =     1600  =     25600 

l'  =  20  =  2000  =  32000  =  512000 
SCALE— ascending,  16,  16,  100,  20  ;  descending,  20,  100,  16,  16. 

NOTE.  The  long  or  gross  ton,  hundred  weight,  and  quarter  were 
formerly  in  common  use ;  but  they  are  now  seldom  used  except  in 
estimating  English  goods  at  the  U.  S.  custom-houses,  and  in  freighting 
and  wholesaling  coal  from  the  Pennsylvania  mines. 

LONG   TON    TABLE. 

28  Ib.  make  1  quarter,  marked    .qr. 

4  qr.  =  112  Ib.  "      1  hundred  weight,        "        cwt. 

20  cwt.  =  2240  Ib.        «      1  ton,  »  T. 

SCALE  —  ascending,  28,  4,  20;  descending,  20,  4,  28. 

The  following  denominations  are  also  in  use. 
56  pounds  make  1  firkin  of  butter. 


100 

« 

1  quintal  of  dried  salt  fish. 

100        " 

M 

1  cask  of  raisins. 

196 

u 

1  barrel  of  flour. 

200        " 

« 

1 

< 

beef,  pork,  or  fish. 

280        " 

« 

1 

< 

salt  at  the  N.  Y.   State 

salt  works. 

56        " 

« 

1  bu 

shel 

a        « 

32        " 

M 

1 

< 

oats. 

48 

" 

1 

i 

barley. 

56 

M 

1 

1      "   corn  or  rye. 

60 

(« 

1        «       "  wheat. 

Define  avoirdupois  weight.  Repeat  the  table.  Give  the  scale.  The 
long  ton  table.  What  other  denominations  are  in  use  ?  "What  is  the 
value  of  each  ? 


164  REDUCTION. 


EXAMPLES    FOR    PRACTICE. 


1.   In  25  T.  15  cwt.  70  Ib. 
how  many  pounds  ? 


OPERATION. 


25  T.  15  cwt.  70  Ib. 


20 

.5 

100 


2.    In   51570   pounds  how 
many  tons  ? 


OPERATION. 


100)  51570  Ib. 


2|0)51|5cwt.  +  701b. 

515  CWt.  OK  m      I     i  r 

25  P. -(-15  cwt. 
Ans.   25  T.  15  cwt.  70  Ib. 


51570  Ib.,  Ans. 

3.  Reduce  3  T.  14  cwt.  74  Ib.  12  oz.  15  dr.  to  drams. 

4.  Reduce  1913551  drams  to  tons. 

5.  A  tobacconist  bought  3  T.  15  cwt.  20  Ib.  of  tobacco,  at 
22  cents  a  pound;  how  much  did  it  cost  him  ?  Ans.  $1654.40. 

6.  How  much  will  115  pounds  of  hay  cost,  at  $10  per  ton  ? 

7.  A  grocer  bought   10  barrels  of  sugar,  each  weighing 

2  cwt.  17  Ib.,  at  6  cents  a  pound;  5  barrels,  each  weighing 

3  cwt.  6  Ib.,  at  7£  cents  a  pound ;  he  sold  the  whole  at  an 
average  price  of  8  cents  a  pound ;  how  much  was  his  whole 
gain?  Ans.    $51.05. 

8.  Paid  $360  for  2  tons  of  cheese,  and  retailed  it  for  12£ 
cents  a  pound  ;  how  much  was  my  whole  gain  ?  Ans.   $140. 

9.  If  a  person  buy  10  T.  6  cwt.  3  qr.  14  Ib.  of  English  iron, 
by  the  long  ton  weight,  at  6  cents  a  pound,  and  sell  the  same 
at  $130  per  short  ton,  how  much  will  he  gain  ?      Ans.    $115.85. 

10.  A  farmer  sold  2  loads  of  corn,  weighing  2352  Ibs.  each, 
at  $.90  per  bu. ;  what  did  he  receive  ?  Ans.  $75.60. 

11.  How  many  pounds  in  300  barrels  of  flour  ?   Ans.  58800. 

12.  A  grocer  bought  3  barrels  of  salt  at  $1.25  per  barrel, 
and  retailed  it  at  J  of  a  cent  per  pound  ?  what  did  he  gain  ? 

Ans.   $2.55. 

STANDARD    OF    WEIGHT. 

188.  In  the  year  1834  the  U.  S.  government  adopted  a 
uniform  standard  of  weights  and  measures,  for  the  use  of  the 
custom  houses,  and  the  other  branches  of  business  connected  with 
the  general  government.  Most  of  the  States  which  have  adopt- 
ed any  standards  have  taken  those  of  the  general  government. 


COMPOUND  NUMBERS.  1G5 

189.  The   United  States  standard  unit  of  weight  is  the 
Troy  pound  of  the  mint,  which  is  the  same  as  the  imperial 
standard  pound  of   Great  Britain,  and  is  determined  as  fol- 
lows :     A  <3ubic  inch  of  distilled  water  in  a  vacuum,  weighed 
by  brass  weights,  also  in  a  vacuum,  at  a  temperature  of  62° 
Fahrenheit's  thermometer,  is  equal  to  252.724  grains,  of  which 
the  standard  Troy  pound  contains  5760. 

190.  The  U.  S.  Avoirdupois  pound  is  determined  from 
the  standard  Troy  .pound,  and  contains   7000  Troy  grains. 
Hence,  the  Troy  pound   is  f  £§ $  =  |Ai  Of  an   avoirdupois 
pound.     But  the  Troy  ounce  contains  -^f^  =  480  grains,  and 
the  avoirdupois  ounce  ^-^p  ==  437.5  grains ;  and  an  ounce  Troy 
is  480 —  437.5  =.  42.5  grains  greater  than  an   ounce   avoirdu- 
pois.    The   pound,   ounce,   and  grain,    Apothecaries'  weight, 
are  the  same  as  the  like  denominations  in  Troy  weight,  the  only 
difference  in  the  two  tables  being  in  the  divisions  of  the  ounce. 

191.  COMPARATIVE   TABLE   OF  WEIGHTS. 

Troy.  Apothecaries'.  Avoirdupois. 

1  pound  =  5760  grains,  =  5760  grains,  =  7000  grains. 
1  ounce  =    480       "       =    480       "       —437.5     " 

175  pounds,  —    175  pounds,  =  144  pounds. 

EXAMPLES    FOR   PRACTICE. 

1.  An    apothecary   bought   5  Ib.   10  oz.  of  rhubarb,  by 
avoirdupois  weight,  at  50  cents  an  ounce,  and  retailed  it  at 
12  cents  a  dram  apothecaries'  weight ;  how  much  did  he  gain? 

Ans.    $33.75. 

2.  Change  424  drams  apothecaries'  weight  to  Troy  weight. 

Ans.    4  Ib.  5  oz. 

3.  Change  20  Ib.  8  oz.  12  pwt.  Troy  weight  to  avoirdu- 
pois weight.  Ans.    17/73-  Ib. 

4.  Bought  by  avoirdupois  weight  20  Ib.  of  opium,  at  40 
cents  an  ounce,  and  sold  the  same  by  Troy  weight  at  50  cents 
an  ounce;  how  much  was  gained  or  lost?        Ans.  $17.83^. 

What  is  the  II.  S.  standard  of  weight  ?  How  obtained  ?  How  is 
the  avoirdupois  pound  determined  ?  How  is  the  apothecaries'  pound 
determined  ?  What  are  the  values  of  the  denominations  of  Troy,  avoir- 
dupois, and  apothecaries'  weight? 


166  REDUCTION. 


MEASURES   OF  EXTENSION. 

11)2.    Extension  has  three  dimensions  —  length,  breadth, 
and  thickness. 

A  Line  has  only  one  dimension  —  length. 

A  Surface  or  Area  has  two  dimensions — length  and  breadth. 

A  Solid  or  Body  has  three  dimensions — length,  breadth,  and 
thickness. 

I.     LONG  MEASURE. 

193.   Long  Measure,  also  called  Linear  Measure,  is  used 
in  measuring  lines  or  distances. 

TABLE. 

12     inches  (in.)  make  1  foot, ft. 

3     feet  "      1  yard, yd. 

5\  yd.,  or  16^  ft.,  "      1  rod rd. 

40     rods  "      1  furlong,  ....  fur. 

8     furlongs,  or  320  rd.,    "      1  statute  mile,,  .mi. 

UNIT    EQUIVALENTS. 

ft.  in. 

yd.  1  =  12 

rd.  '1      —  3  =  36 

fur.  1  =        51  =       16*  =      198 

mi        1  =    40  —     220"  =     666"  =     7920 
1  =  8  =  320  —  1760    =  5280§    =  63360 
SCALE  — ascending,  12,  3,  5J,  40,  8;  descending,  8,40,  =>£,  3,  J2. 

The  following  denominations  are  also  in  use :  — 

3  barleycorns    make    1    inch,  5  ll,se(|  b?  shoemakers  in  measuring 

'  (  the  length  of  the  foot. 

4  inches  "       1  hand  5  used  in  .measm'hl£   t he   height  of 

'  (  horses  directly  over  the  fore  feet. 
6  feet  "       1  fathom,  used  in  measuring1  depths  at  sou, 

1.15  statute  miles  «      1  geographic  mile,  J  "8Cfl  in  measuring  dis- 

'  ^  tances  at  sea. 
3  geographic   "       "      1  league. 

60         "  «       «   >  1    ,  <  of  latitude   on  a  meridian  or  of 

69|  statute      «       «   ^      Degree  ^  ]ongjtude  on  the  equator. 
S60  degrees  "      the  circumference  of  the  earth. 

TTow  many  dimensions  has  extension?     Define  a  line.     Surf?! <•<>  or 
;in  a.     A  solid  or  body.     Define  long  measure.     What  are  the  d> 

The  value  of  each.     What  other  denominations  arc  us:  d  ? 


COMPOUND    NUMBERS. 


107 


NOTES.  1.  For  the  purpose  of  measuring  cloth  and  other  goods  sold 
by  the  yard,  the  yard  is  divided  into  halves,  fourths,  eighths,  and  six- 
teenths. The  old  table  of  cloth  measure  is  practically  obsolete. 

2.  The  geographic  mile  is  ^V  ofy|?r  °r  Ti^To  °^  ^e  distance  round 
the  center  of  the  earth.     It  is  a  small  fraction  more  than  1.15  statute 
miles. 

3.  The  length  of  a  degree  of  latitude  varies,  being  68.72  miles  at  the 
equator,  68.9  to  69.05  miles  in  middle  latitudes,  and  69.30  to  69.34  miles 
in  the  polar  regions.     The  mean  or  average  length  is  as  stated  in  the 
table.     A  degree  of  longitude  is  greatest  at  the  equator,  where  it  is 
69.16  miles,  and  it  gradually  decreases  toward  the  poles,  where  it  is  0. 


EXAMPLES    FOR   PRACTICE. 


1.    In  2  mi.  4  fur.  32  rd. 
2  yd.  how  many  inches  ? 

OPERATION. 

2  mi.  4  fur.  32  rd.  2yd. 
8 


20fiir. 
40 

~832rd. 

5* 
416 
4162 

4578yd. 
3 


13734ft. 
12 


2.    In   164808   inches  how 
many  miles? 

OPERATION. 
12)  164808  in. 
3)13734  ft. 


54)4578  yd. 
_2j        2 

Tl  ) 9156 

4|0)83|2rd.  +  f  yd.  =  2  yd. 
8)20fur.-f-32rd. 
~~2  mi.  +  4  fur. 
Ans.   2  mi.  4  fur.  32  rd.  2  yd. 


164808  in.,  Ans. 

3.  The  diameter  of  the  earth  being  7912  miles,  how  many 
inches  is  it?  Ans.    501304320  inches. 

4.  In  168474  feet  how  many  miles  ? 

5.  In  31  mi.  7  fur.  10  rd.  3  yd.,  how  many  feet  ? 

6.  If  the  greatest  depth  of  the  Atlantic  telegraphic  cable 
from  Newfoundland  to  Ireland  be  2500   fathoms,  how  many 
miles  is  it?  Ans.    2  mi.  6  fur.  29  rd.  1    ft. 


168 


REDUCTION. 


7.  If  this  cable  be  2200  miles  in  length,  and  cost  10  cents 
a  foot,  what  was  its  whole  cost?  Ans.   $1161600. 

8.  A  pond  of  water  measures  4  fathoms  3  feet  8  inches  in 
depth  ;  how  many  inches  deep  is  it  ?  Ans.    332. 

9.  How  many  times  will  the  driving  wheels  of  a  locomo- 
tive turn  round  in  going  from  Albany  to  Boston,  a  distance  of 
200  miles,  supposing  the  wheels  to  be  18  ft.  4  inches  in  cir- 
cumference? Ans.    57600  times. 

10.  If  a  vessel  sail  120  leagues  in  a  day,  how  many  stat- 
ute miles  does  she  sail  ?  Ans.    414. 

11.  How  many  inches  high  is  a  horse  that  measures  14J- 
hands?  Ans.    58. 

SURVEYORS'  LONG  MEASURE. 

194.   A  Gunter's  Chain,  used  by  land  surveyors,  is  4  rods 
or  66  feet  long,  and  consists  of  100  links. 

TABLE. 

* 

7.92  inches  (in.)  make  1  link, 1. 

25  links  "  1  rod, rd. 

4  rods,  or  66  feet,  "  1  chain.,  ch. 

80  chains  "  1  mile,  . .  mi. 

UNIT    EQUIVALENTS. 

1.  in. 

rd.  1   =       7.92 

ch.  1  =       25  i=       198 

mi.          1  —       4  =     100  =       792 
1  =  80  =  320  =  8000  =  63360 

SCALE  — ascending,  7.92,  25,  4,  80  ;  descending,  80,  4,  25,  7.92. 

NOTE.     Rods  are  seldom  used  in  chain  measure,  distances  being 
taken  in  chains  and  links.    „ 

EXAMPLES    FOR    PRACTICE. 

1.  In  3  mi.  51  ch.  73  1.  how  many  links? 

2.  Reduce  29173  1.  to  miles. 

3.  A  certain  field,  enclosed  by  a  board  fence,  is  17  ch.  31 1. 
long,  and  12  ch.  87  1.  wide  ;  how  many  feet  long  is  the  fence 
which  encloses  it  ?  Ans.    3983.76ft. 


Repeat  the  table  of  surveyors'  long  measure.     Give  the  scale. 


COMPOUND   NUMBERS. 


169 


12  in. 


1ft. 


II.     SQUARE  MEASURE. 

A  Square  is  a  figure  having  four  equal  sides,  and 
four  equal  angles  or  corners. 

1  square  foot  is  a  figure  having  four 
sides  of  1  ft.  or  12  in.  each,  as  shown 
in  the  diagram.  Its  contents  are  12 
X  12  ^=:  144  square  inches.  Hence 

The  contents  or  area  of  a  square,  or 
of  any  other  figure  having  a  uniform 
length  and  a  uniform  breadth,  is  found 
by  multiplying  the  length  by  the  breadth. 
Thus,  a  square  foot  is  12  in.  long  and  12  in.  wide,  and  the  con- 
tents are  12  X  12  =  144  square  inches.  A  board  20  in.  long 
and  10  in.  wide,  is  a  rectangle,  containing  20  X  10  =  200 
square  inches. 

1OO .    Square  Measure  is  used  in  computing  areas  or  sur- 
faces ;  as  of  land,  boards,  painting,  plastering,  paving,  &c. 

TABLE. 

144    square  inches  (sq.  in.)  make  1  square  foot,  marked  sq.  ft. 
9    square  feet       •  1  square  vard,       "      so.  vd. 


12  in.  =  1  ft. 


y    square  feet       • 
301  square  yards 
40    square  rods 

4    roods 
640    acres 


1  7 

square  yard, 
i  square  rod, 
1  rood, 
1  acre, 
1  square  mile, 


R. 


UNIT   EQUIVALENTS 


sq.  yd. 

1  = 

30*-  = 
1210  = 


272|  = 
10390  = 


sq.  yd. 
sq.  rcl. 

R. 

A. 

"      sq.  mi. 


sq. in. 

144 
1296 
39204 
1568 160 


sq.mi.     1=       4=        160=        4840=        43560=        6272640 
1  =  640  =  2560  =  102400  =  3097600  =  27878400  =  4014489600 
SCALE  — ascending,  144,  9,  301,  40,  4,  640;  descending,  640,  4, 
40,  301,  9,  144. 


Define  a  square.  How  is  the  area  of  a  square  or  any  rectangular 
figure  found  ?  For  what  is  square  measure  used  ?  Repeat  the  table. 
Give  the  scale. 


170  REDUCTION. 

Artificers  estimate  their  work  as  follows : 

By  the  square  foot :  glazing  and  stone-cutting. 

By  the  square  yard :  painting,  plastering,  paving,  ceiling,  and 
paper-hanging. 

By  the  square  of  100  feet:  flooring,  partitioning,  roofing,  slating, 
and  tiling. 

Brick-laying  is  estimated  by  the  thousand  bricks;  also  by  the 
square  yard,  and  the  square  of  100  feet. 

NOTES.  1.  In  estimating  the  painting  of  moldings,  cornices,  &c.,  the 
measuring-line  is  carried  into  all  the  moldings  and  cornices. 

2.  In  estimating  brick-laying  by  the  square  yard  or  the  square  of 
100  feet,  the  work  is  understood  to  be  1-^  bricks,  or  12  inches,  thick. 

EXAMPLES    FOR    PRACTICE. 

1.  In  10  A.  1  R.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 
how  many  square  inches  ? 

OPERATION. 

10  A.  1  R.  25  sq.  rd.  16  sq.yd.  4  sq.  ft.  136  sq.  in. 

4 

41  R. 
40 


1665sq.rd. 
30j 

416J 
49966 


50382J  sq.  yd. 
_  9_ 

453444i  sq.ft. 
144 


36  =  £  sq.  ft, 
1813912  with  136  sq.in. 
1813776 
453444 

65296108  sq.  in.,  Ans. 
2.   In  65296108  sq.  in.  how  many  acres? 

How  do  artisans  estimate  work  ? 


COMPOUND  NUMBERS.  171 

OPERATION. 

144  )  65296108  sq.  in. 

9  )  453445  sq.  ft.  +  28  sq.  in. 
30f  l  50382  sq.  yd.  +  7  sq.  ft. 


121  )  201 528  fourths  sq.  yd. 

4|0)166|5  sq.  rd.  +  ^=15}  sq.  yd. 
4)41R.-f-25sq.rd. 

10  A.  +  1  R. 

Ans.   10  A.  1  R.  25  sq.  rd.  lof  sq.  yd.  7  sq.  ft.  28  sq.  in. 

C  10  A.  1  R.  25  sq.  rd.  15  sq.  yd.  7  sq.  ft.  28  sq.  in. 

Or  3  6  sq.  ft.  108  sq.  in. 

Or       10  A.  1  R.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 

ANALYSIS.  Dividing  by  the  numbers  in  the  ascending  scale,  and 
arranging  the  remainders  according  to  their  order  in  a  line  below, 
we  find  the  square  yards  a  mixed  number,  15-|.  But  %  of  a  sq.  yd. 
—  |  of  9  sq.  ft.  =  6|  sq.  ft.  ;  and  f  of  a  sq.  ft.  =  f  of  144  sq.  in.  = 
108  sq.  in.  Therefore  ^  sq.  yd.  :=  6  sq.  ft.  108  sq.  in. ;  and  adding 
108  sq.in.  to  28  sq.  in.  we  have  136  sq.  in.,  and  6  sq.  ft.  to  7  sq.ft.  we 
have  13  sq.  ft.  =  1  sq.  yd.  4  sq.  ft.,  and  writing  the  4  sq.  ft.  in  the 
result,  and  adding  1  sq.  yd.  to  15  sq.  yd.  we  have  for  the  reduced 
result,  10  A.  1  R.  25  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 

3.  Reduce  87  A.  2  R.  38  sq.  rd.  7  sq.  yd.  1  sq.  ft.  100  sq. 
in.  to  square  inches.  4ns.    550355068  sq.  in. 

4.  Reduce  550355068  square  inches  to  acres. 

5.  A  field  100  rods  long   and  30  rods  wide  contains  how 
many  acres  ?  Ans.    18  A.  3  R. 

6.  How  many  rods  of  fence  will   enclose  a  farm  a  mile 
square?  4ns.    1280  rods. 

7.  How  much  additional  fence  will  divide  it  into  four  equal 
square  fields  ?  4ns.    640  rods. 

8.  How  many  acres  of  land  in  Boston,  at  $1  a  square  foot, 
will  $100000  purchase  ? 

Ans.   2  A.  1  R.  7  sq.  rd.  9  sq.  yd.  3J  s.q.  ft. 

9.  How  many  yards  of  carpeting,  1  yd.  wide,  will  be  required 
to  carpet  a  room  18£  ft.  long  and  16  ft.  wide  ?  4ns.  32|  yd. 


172  REDUCTION. 

10.  What  would  be  the  cost  of  plastering  a  room  18  ft.  long, 
16£  ft.  wide,  and  9  ft.  high,  at  22  cts.  a  sq.  yd.  ?  Ans.  $22.44. 

11.  What  will  be  the  expense  of  .slating  a  roof  40  feet 
long  and  each   of  the  two    sides   20   feet  wide,  at    $10  per 
square?  Ans.    $160. 

SURVEYORS'  SQUARE  MEASURE. 

1OT.    This  measure  is  used  by  surveyors  in  computing  the 
area  or  contents  of  land. 

TABLE. 

625  square  links  (sq.  1.)  make  1  pole, P. 

16  poles  "      1  square  chain,,  .sq.  ch. 

10  square  chains  "      1  acre, A. 

640  acres  "      1  square  mile,.  ..sq.  mi. 

36  square  miles  (6  miles  square)     "      1  township, Tp. 

UNIT    EQUIVALENTS. 

P.  sq.  1. 

sq.  Ch.  1  =  625 

A.  1  =  16  ~  1000 

sq.  mi.  1  =  10  =  160  —  10000 

Tp.        1  =      640  =      6100  =    102400  =.      64000000 
1  =  36  =  23040  =.  230400  =  3686400  =  2304000000 
SCALE  —  ascending,  625,  16,  10,  640,  36;   descending,  36,  640, 
10,  16,  625. 

NOTES.     1.  A  square  mile  of  land  is  also  called  a  section. 

2.  Canal  and  railroad  engineers  commonly  use  an  engineers'  chain, 
which  consists  of  100  links,  each  1  foot  long. 

3.  The  contents  of  land  are  commonly  estimated  in  square  miles, 
acres,  and  hundredths  ;  the  denomination,  rood,  is  fast  going  into  dis- 
use. 

EXAMPLES    FOR    PRACTICE. 

1.  How  many  poles  in  a  township  of  land  ? 

2.  Reduce  3686400  P.  to  sq.  mi. 

3.  In  94  A.  7  sq.  ch.  12  P.   118  sq.  1.  how  many  square 
links? 

4.  What  will  be  the  cost  of  a  farm  containing   4550000 
square  links,  at  $50  per  acre  ?  Ans.     $2275. 

Repeat  the  table  of  surveyors'  square  measure.     Give  the  scale. 


COMPOUND   IS  UMBERS.  173 


III.     CUBIC  MEASURE. 

198.  A  Cube  is  a  solid,  or  body, 
having  six  equal  square  sides,  or 
faces.  If  each  side  of  a  cube  be  1 
yard,  or  3  feet,  1  foot  in  thickness 
of  this  cube  will  contain  3X3X1 
=±  9  cubic  feet,  and  the  whole  cube  will 
contain  3  X  3  X  3  —  27  cubic  feet. 

olt.  =1  yd.  O 

A  solid,  or  body,  may  have  the 

three  dimensions  all  alike  or  all  different.  A  body  4  ft.  long, 
3  ft.  wide,  and  2  ft,  thick  contains  4  X  3  X  2  =  24  cubic  or 
solid  feet.  Hence  we  see  that 

The  cubic  or  solid  contents  of  a  body  are  found  by  multiply- 
ing the  length,  breadth,  and  thickness  together. 

199.  Cubic  Measure,  also  called  Solid  Measure,  is  used 
in  estimating  the  contents  of  solids,  or  bodies ;  as  timber,  wood, 
stone,  &c. 

TABLE. 

1728  cubic  inches  (cu.  in.)  make  1  cubic  foot, cu.  ft. 

27  cubic  feet  "  1  cubic  yard, cu.  yd. 

16  cubic  feet  "  1  cord  foot, cd.  ft. 

8  cord  feet,  or  ?  (t  l        d     f         d        ^ 
128  cubic  feet,     $ 

rt/l  „,./,,  ..  -  (  perch  of  stone  ?  -^  , 

24}  cubic  feet  1  |  g  masonrV)      \  Pch. 

SCALE  —  ascending,  1728,  27.  The  other  numbers  are  not  in  a 
regular  scale,  but  are  merely  so  many  times  1  foot.  The  unit 
equivalents,  being  fractional,  are  consequently  omitted. 

NOTES.     1.  A  cubic  yard  of  earth  is  called  a  load. 

2.  Railroad  and  transportation  companies  estimate  light  freight  by 
the  space  it  occupies  in  cubic  feet,  and  heavy  freight  by  weight. 

3.  A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains 
1  cord ;  and  a  cord  foot  is  1  foot  in  length  of  such  a  pile. 

4.  A  perch  of  stone  or  of  masonry  is  16£  feet  long,  1^  feet  wide,  and 
1  foot  high. 

Define  a  cube.  How  are  the  contents  of  a  cube  or  rectangular 
solid  found?  For  what  is  cubic  measure  used?  Repeat  the  table. 
Give  the  scale.  How  is  railroad  freight  estimated  ?  What  is  under- 
stood by  a  cord  foot  ?  By  a  perch  of  stone  or  masonry  ? 


174  REDUCTION. 

5.  Joiners,  bricklayers,  and  masons  make  no  allowance  for  windows, 
doors,  &c.      Bricklayers  and  masons,  in  estimating   their  work   by 
cubic  measure,  make  no  allowance  for  the  corners    of  the  walls  of 
houses,  cellars,  &c.,  but  estimate  their  work  by  the  girt,   that  is,  the 
entire  length  of  the  wall  on  the  outside. 

6.  Engineers,  in  making  estimates  for  excavations  and  embankments, 
take  the  dimensions  with  a  line  or  measure  divided  into  feet  and  deci- 
mals of  a  foot.     The  estimates  are  made  in  feet  and  decimals,  and  the 
results  are  reduced  to  cubic  yards. 

EXAMPLES    FOR    PRACTICE. 

1.  In  125  cu.  ft.  840  cu.  in.  how  many  cu.  in.  ?  Ans.  216840. 

2.  Reduce  5224  cubic  feet  to  cords.  Ans.   40|f. 

3.  In  a  solid,  3  ft.  2  in.  long,  2  ft.  2  in.  wide,  and  1  ft.  8  in. 
thick,  how  many  cubic  inches?  Ans.    19760. 

4.  How  many  small  cubes,  1  inch  on  each  edge,  can  be 
sawed  from  a  cube  6  feet  on  each  edge,  allowing  no  waste  for 
sawing?  Ans.    373248. 

5.  In  a  pile  of  wood  60  feet  long,  20  feet  wide,  and  15  feet 
high,  how  many  cords  ?  Ans.    140-|. 

6.  How  many  cubic  feet  in  a  load  of  wood  10  feet  long,  3| 
feet  wide,  and  3£  feet  high  ?  Ans.    113|  cu.  ft. 

7.  If  a  load  of  wood  be  12  feet  long  and  3  feet  wide,  how 
high  must  it  be  to  make  a  cord  ?  Ans.    3f  ft.  high. 

8.  The  gray  limestone  of  Central  New  York  weighs  175 
pounds  a  cubic  foot.     What  is  the  weight  of  one  solid  yard  ? 

Ans.   2  T.  7  cwt.  25  Ib. 

9.  A  cellar  wall,  32  ft.  by  24  ft.,  is  6  ft,  high  and  1£  ft.  thick. 
How  much  did  it  cost  at  $1.25  a  perch?      Ans.    $50.909+ 

10.  How  much  did  it  cost  to  dig  the  same  cellar,  at  15 
cents  a  cubic  yard  ?  AM.    $25.60. 

11.  My  sleeping  room  is  10  ft.  long,  9  ft.  wide,  and  8  ft.  high. 
If  I  breathe  10  cu.  ft.  of  air  in  one  minute,  in  how  long  a  time  will 
I  breathe  as  much  air  as  the  room  contains  ?       Ans.    72  min. 

12.  In  a  school  room  30  ft.  long,  20  ft.  wide,  and  10  ft.  high, 
with  50  persons  breathing  each  10  cu.  ft,  of  air  in  one  minute, 
in  how  long  a  time  will  they  breathe  as  much  as  the  room 
contains?  Ans.    12  min. 

How  are  excavations  and  embankments  measured  ? 


COMPOUND    NUMBERS.  175 


MEASURES   OF   CAPACITY. 

I.     LIQUID  MEASURE. 

194.  Liquid  Measure,  also  called  Wine  Measure,  is  used 
in  measuring  liquids  ;  as  liquors,  molasses,  water,  &c. 

TABLE. 

4    gills  (gi.)  make  1  pint, pt. 

2    pints  "       1  quart, qt. 

4    quarts  "      1  gallon, gal. 

31J  gallons  «      1  barrel,  . ..  ..bbl. 

2    barrels,  or  63  gal.     "      1  hogshead,,  .hhd. 

UNIT   EQUIVALENTS. 

pt.  gi. 

qt.  1    —  4 

gal.  1=2=  8 

bbi.         1    —      4  =      8  =      32 

hhd.       1  =  3U  —  126  =  252  =  1008 

1  =  2  =  63    =  252  —  504  =  2016 

SCALE  —  ascending,  4,  2,  4,  31£,  2 ;  descending,  2,  31£,  4,  2, 4. 

The  following  denominations  are  also  in  use : 

36  gallons  make  1  barrel       of  beer. 

54      "        or  1^  barrels  "     1  hogshead  "     " 

42       "  "     1  tierce. 

2  hogsheads,  or  120  gallons,  "     1  pipe  or  butt. 

2  pipes  or  4  hogsheads,  "     1  tun. 

NOTES.  1.  The  denominations,  barrel  and  hogshead,  are  used  in  es- 
timating the  capacity  of  cisterns,  reservoirs,  vats,  &c. 

2.  The  tierce,  hogshead,  pipe,  butt,  and  tun  are  the  names  of  casks, 
and  do  not  express  any  fixed  or  definite  measures.     They  are  usually 
gauged,  and  have  their  capacities  in  gallons  marked  on  them. 

3.  Ale  or  beer  measure,  formerly  used  in  measuring  beer,  ale,  and 
milk,  is  almost  entirely  discarded. 

What  is  liquid  measure  ?  Repeat  the  table.  Give  the  scale.  What 
other  denominations  are  sometimes  used  ?  How  are  the  capacities  of 
cisterns,  reservoirs,  &c.,  reckoned  ?  Of  large  casks  ? 


176 


EEDUCTION. 


EXAMPLES    FOR    PRACTICE. 


1.  In  2  hhd.  1  bar.  30  gal.  2 
qt.  1  pt.  3  gi.  how  many  gills  ? 

OPERATION. 
2  hhd.  1  bar.  30  gal.  2  qt. 


2 

5bbl. 


[lpt.3gi. 


185 

187^  gal. 
4 

752  qt. 
_2 

1505pt. 
4 


2.    In  6023  gi.  how  many 
hhds.  ? 

OPERATION. 

4  )  6023  gi. 
2  )  1505  pt.  +  3  gi. 
4  )  752  qt.  +  1  pt. 


188  gal. 

j      2 


63   )376 


[gal. 


2  hhd.  +  1  bar. 
Ans.    2  hhd.  1  bar.  30£  gal. 

1  pt.  3  gi. 

But  J-  gal.  =  2  qt.,  making 
the  Ans.  2  hhd.  1  bar.  30  gal. 

2  qt.  1  pt.  3  gi. 


6023  gi.,  Ans. 

3.  Reduce  3  hogsheads  to  gills. 

4.  Reduce  6048  gills  to  hogsheads. 

5.  In  13  hhd.  15  gal.  1  qt.  how  many  pints? 
6..    In  6674  pints  how  many  hogsheads  ? 

7.  What  will  be  the  cost  of  a  hogshead  of  wine,  at  6  cents 
a  gill?  Ans.    $120.96. 

8.  A  grocer  bought   10  barrels  of  cider,  at  $2  a  barrel; 
after  converting  it  into  vinegar,  he  retailed  it  all  at  5  cents  a 
quart ;  how  much  was  his  whole  gain  ?  Ans.    $43. 

9.  At  6  cents  a  pint,  how  much  molasses  can  be  bought  for 
$3.84?  Ans.  8  gal. 

10.  How  many  demijohns,  that  will  contain  2  gal.  2  qt.  1  pt. 
each,  can  be  filled  from  a  hogshead  of  wine  ?          Ans.   24. 

II.     DRY  MEASURE. 

UOo.    Dry  Measure    is    used   in   measuring  articles   not 
liquid,  as  grain,  fruit,  salt,  roots,  ashes,  &c. 


What  is  dry  measure  ? 


COMPOUND  NUMBERS.  177 


TABLE. 

2  pints  (pt.)  make  1  quart,  ..........  qt. 

8  quarts  "       1  peck,  .........  pk. 

4  pecks  "       1  bushel,.  bu.  or  bush. 

UNIT    EQUIVALENTS. 
qt.  pt. 

pk.  1  =  2 
bu.  1  =  8  =  16 
1  =  4  =  3£  =  64 

SCALE  —  ascending,  2,  8,  4  ;   descending,  4,  8,  2. 

NOTE.  In  England,  8  bu.  of  70  Ibs.  each  are  called  a  quarter,  used  in 
measuring  grain.  The  weight  of  the  English  quarter  is  •£•  of  a  long  ton. 

EXAMPLES    FOR    PRACTICE. 

1.  In  49  bu.  3  pk.  7  qt.  1  pt.  how  many  pints  ? 

2.  In  3199  pt.  how  many  bushels  ? 

3.  Reduce  1  bu.  1  pk.  1  qt.  1  pt.  to  pints. 

4.  Reduce  83  pints  to  bushels. 

5.  An  innkeeper  bought  a  load  of  50  bushels  of  oats  at  65 
cents  a  bushel,  and  retailed  them  at  25  cents  a  peck  ;  how 
much  did  he  make  on  the  load  ?  Ans.   $17.50. 

STANDARD    OF    EXTENSION. 

2O2*  The  U.  S.  standard  unit  of  measures  of  extension, 
whether  linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or  36 
inches,  and  is  the  same  as  the  imperial  standard  yard  of 
Great  Britain.  It  is  determined  as  follows  :  The  rod  of  a 
pendulum  vibrating  seconds  of  mean  time,  in  the  latitude  of 
London,  in  a  vacuum,  at  the  level  of  the  sea,  is  divided  into 
391393  equal  parts,  and  360000  of  these  parts  are  36  inches, 
or  1  standard  yard.  Hence,  such  a  pendulum  rod  is  39.1393 
inches  long,  and  the  standard  yard  is  f  f  ^$$$  of  the  length  of 
the  pendulum  rod. 

2O  3.  The  U.  S.  standard  unit  of  liquid  measure  is  the  old 
English  wine  gallon,  of  231  cubic  inches,  which  is  equal  to 
8.339  pounds  avoirdupois  of  distilled  water  at  its  maximum 
density,  that  is,  at  the  temperature  of  39.83°  Fahrenheit,  the 
barometer  at  30  inches. 

Repeat  the  table.  What  is  a  quarter  ?  What  is  the  U.  S.  standard 
unit  of  measurement  of  extension  ?  How  is  it  determined  ?  What  is 
the  U.S.  standard  unit  of  liquid  measure  ? 


178  REDUCTION. 

2O  4.  The  U.  S.  standard  unit  of  dry  measure  is  the  Brit- 
ish Winchester  bushel,  which  is  18£  inches  in  diameter  and  8 
inches  deep,  and  contains  2150.42  cubic  inches,  equal  to 
77.6274  pounds  avoirdupois  of  distilled  water,  at  its  maximum 
density.  A  gallon,  dry  measure,  contains  268.8  cubic  inches. 

NOTE.  1.  The  wine  and  dry  measures  of  the  same  denomination 
are  of  different  capacities.  The  exact  and  the  relative  size  of  each  may 
be  readily  seen  by  the  following 

2O5.  COMPARATIVE  TABLE  OF  MEASURES  OF  CAPACITY. 

Cu.  in.  in        Cu.  in.  in         Cu.  in.  in          Cu.  in.  in 
one  gallon.      one  quart.       one  pint.  one  gill. 

Wine  measure,  231  57|  28J  7, 

Dry  measure,  (\  pk.,)  268f          67j  33-f 

2.  The  beer  gallon  of  282  inches  is  retained  in  use  in  a  few  places 
only  by  custom. 

EXAMPLES    FOR   PRACTICE. 

1.  A   fruit   dealer  bought  a  bushel  of  strawberries,  dry 
measure,  and  sold  them  by  wine  measure  ;  how  many  quarts 
did  he  gain  ?  Am.    5£f  quarts. 

2.  A  grocer  bought  40  quarts  of  milk  by  beer  measure,  and 
sold  it  by  wine  measure ;  how  many  quarts  did  he  gain  ? 

Ans.   8f  |-  quarts. 

3.  A  bushel,  or  32  quarts,  dry  measure,  contains  how  many 
more  cubic  inches  than  32  quarts  wine  measure  ? 

Ans.    302f  cu.  in. 
TIME. 

2O6.  Time  is  used  in  measuring  periods  of  duration,  as 
years,  days,  minutes,  &c. 

TABLE. 

60  seconds  (sec.)  make  1  minute, min. 

60  minutes  "      1  hour, h. 


24  hours 
7  days 

365  days 

366  days 

12  calendar  months 


1  day, da. 

1  week, wk. 

1  common  year,. .  .yr. 

1  leap  year, yr. 

1  year, yr. 


100  years  "      1  century, C. 


What  is  the  IT.  S.  standard  unit  of  dry  measure  ?  How  is  it  ob- 
taiiK  d  r  'What  is  tlu:  relative  si/o  of  the  wine  and  the  dry  gallon? 
What  is  the  ,si/e  of  a  beer  gallon  ?  What  is  time  ?  Repeat  the  table. 


COMPOUND   NUMBERS. 


179 


wk. 
1  : 

yr.        mo. 
1  =  12  : 


UNIT   EQUIVALENTS. 

min.  sec. 

b.                         1  =                    60 

da.                   1    =              60  —              3600 

I  —      24  =       1440  =        86400 

7  =     168  =     10080  =      604800 

365  =  8760  =  525600  =  31536000 

366  =i  8784  =  527040  =  31622400 


SCALE  —  ascending,  60,  60,  24,  7,  4;  descending,  4,  7,  24,  60,  60. 
The  calendar  year  is  divided  as  follows :  — 


No.  of  mo. 

1 

2 

3 

4 

5 

6 

7 

8 

9 
10 
11 
12 


Season. 

Winter, 

« 

Spring, 


Summer, 
u 


Autumn, 


Winter, 


Names. 

Abbreviations. 

January, 

Jan. 

February, 

Feb. 

March, 

Mar. 

April, 

Apr. 

rfay, 

June, 

Jun. 

July, 



August, 

Aug. 

(T  September, 

Sept. 

<  October, 

Oct. 

d  November, 

Nov. 

December, 

Dec. 

No.  of  days. 

31 

28  or  29 

31 

30 

31 

30 

31 

31 

30 

31 

30 

31 


365  or  366 

NOTES.  1.  The  exact  length  of  a  solar  year  is  365  da.  5  h.  48  min.  46 
sec. ;  but  for  convenience  it  is  reckoned  11  min.  14  sec.  more  than  this, 
or  365  da.  6  h.  =  365^  da.  This  ^  day  in  4  years  makes  one  day, 
which,  every  fourth,  bissextile,  or  leap  year,  is  added  to  the  shortest 
month,  giving  it  29  days.  The  leap  years  are  exactly  divisible  by  4, 
as  1856,  1860,  1864.  The  number  of  days  in  each  calendar  month 
may  be  easily  remembered  by  committing  the  following  lines  :  — 

"  Thirty  days  hath  September, 
April,  June,  and  November ; 
All  the  rest  have  thirty-one, 
Save  February,  which  alone 
Hath  twenty-eight;  and  one  day  more 
We  add  to  it  one  year  in  four." 

2.  In  most  business  transactions  30  days  are  called  1  month. 
EXAMPLES    FOR    PRACTICE. 

1.  Reduce  365  da.  5  h.  48  min.  46  sec.  to  seconds. 

2.  Reduce  31556926  seconds  to  days. 


Give  the  scale.  What  is  the  length  of  each  of  the  calendar  months  ? 
What  is  the  exact  length  of  a  solar  year  r  Explain  the  use  of  bissextile 
or  leap  year.  What  is  the  length  of  a  month  in  business  transactions  ? 


180  REDUCTION. 

3.  In  5  wk.  1  da.  1  h.  1  min.  1  sec.  how  many  seconds  ? 

4.  In  3114061  seconds  how  many  weeks? 

5.  How  many  times  does  a  clock  pendulum,  3  ft.  3  in.  long, 
beating  seconds,  vibrate  in  one  day?  Ans.    86400. 

6.  If  a  man  take  1  step  a  yard  long  in  a  second,  in  how 
lono-  a  time  will  he  walk  10  miles  ?  Ans.  4  h.  53  min.  20  sec. 

O 

7.  In  a  lunar  month  of  29  da.  12  h.  44  min.  3  sec.  how 
many  seconds?  Ans.   2551443. 

8.  How  much  time  will  a  person  gain  in  40  years,  by  rising 
45  minutes  earlier  every  day  ?      Ans.  456  da.  13  h.  30  min. 

CIRCULAR  MEASURE. 

SOT.  Circular  Measure,  or  Circular  Motion,  is  used  prin- 
cipally in  surveying,  navigation,  astronomy,  and  geography, 
for  reckoning  latitude  and  longitude,  determining  locations  of 
places  and  vessels,  and  computing  difference  of  time. 

Every  circle,  great  or  small,  is  divisible  into  the  same  num- 
ber of  equal  parts,  as  quarters,  called  quadrants,  twelfths, 
called  signs,  360ths,  called  degrees,  &c.  Consequently  the 
parts  of  different  circles,  although  having  the  same  names,  are 
of  different  lengths. 

TABLE. 

60  seconds  (")      make  1  minute, . . . '. 

60  minutes  "       1  degree, . . .  °. 

30  degrees  "       1  sign, S. 

12  signs,  or  360°,    "       1  circle, C. 

UNIT   EQUIVALENTS. 

{    _-  fo 

s.  1  =         GO  =        3600 

C.          1  =    30  =     1800  =     108000 
1  =  12  =  360  =  21600  =  1296000 
SCALE  —  ascending,  60,  60,  30, 12;  descending,  12,  30,  60,  60. 

NOTES.  1.  Minutes  of  the  earth's  circumference  are  called  geo- 
graphic or  nautical  miles. 

2.  The  denomination,  signs,  is  confined  exclusively  to  Astronomy. 

Define  circular  measure.  How  are  circles  divided  ?  Repeat  the 
table.  Give  the  scale.  What  is  a  geographic  mile?  What  is  a 
feign.? 


COMPOUND  NUMBERS.  181 

3.  Degrees  are  not  strictly  divisions  of  a  circle,  but  of  the  space 
about  a  point  in  any  plane. 

4.  90°  make  a  quadrant,  or  right  angle,  and  60°  a  sextant,  or  ^  of  a 
circle. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  10  S.  10°  10'  10"  to  seconds. 

2.  Reduce  1116610''  to  signs. 

3.  How  many  degrees  in  11400   geographic  or   nautical 
miles?  Ans.    190°. 

4.  If  1  degree  of  the  earth's  circumference  is  69£  statute 
miles,  how  many  statute  miles  in  11400  geographic  miles,  or 
190  degrees?  Ans.  13148. 

5.  How  many  minutes,  or  nautical  miles,  in  the   circum- 
ference of  the  earth?  Ans.  21  GOO7  or  mi. 

6.  A  ship  during  4  days'  storm  at  sea  changed  her  longitude 
397  geographical  miles  ;  how  many  degrees  and  minutes  did 
she  change  ?  Ans.   6°  37'. 


IN  COUNTING. 

12  units  or  things.  .  .  .make.  .  .  .1  dozen. 

12  dozen  "  1  gross. 

12  gross  "  1  great  gross. 

20  units  "  1  score. 

3O9.     PAPER. 

24  sheets  ........  make  ........  1  quire. 

20  quires  "  1  ream. 

2  reams  "  1  bundle. 

5  bundles  "  1  bale. 

S1O.      BOOKS. 

The  terms  folio,  quarto,  octavo,    duodecimo,  Sec.,  indicate 
the  number  of  leaves  into  which  a  sheet  of  paper  is  folded. 
A  sheet  folded  in    2  leaves  is  called  a  folio. 


A  sheet  folded  in  4  leaves 
A  sheet  folded  in  8  leaves 
A  sheet  folded  in  12  leaves 
A  sheet  folded  in  16  leaves 
A  sheet  folded  in  1 8  leaves 
A  sheet  folded  in  24  leaves 
A  sheet  folded  in  32  leaves 


a  quarto,  or  4to. 

an  octavo,  or  8vo. 

a  12mo. 

a  16mo. 

an  ISmo. 

a  24mo. 

a  32mo. 


What  is  a  degree  ?     Repeat  the  table  for  counting.     For  reckoning 
paper.    For  indicating  the  size  of  books. 


182  REDUCTION. 

EXAMPLES  FOR  PRACTICE. 

1.  If  in  Birmingham,  England,  150  million  Gillott  pens  are 
manufactured  annually,  how  many  great  gross  will  they  make  ? 

Ans.    86805  great  gross  6  gross  8  dozen. 

2.  In  100000  sheets  of  paper,  how  many  bales  ? 

Ans.    20  bales  4  bundles  6  quires  16  sheets. 

3.  What  is  the  age  of  a  man  4  score  and  10  years  old  ? 

4.  How  many  printed  pages,   2   pages  to  each  leaf,  will 
there  be  in  an  octavo  book,  having  8  fully  printed  sheets  ? 

Ans.    128  pages. 

5.  How  large  a  book  will  ten  32mo.  sheets  make,  if  every 
page  be  printed  ?  Ans.    640  pages. 

PROMISCUOUS    EXAMPLES    IN    REDUCTION. 

1.  How  many  suits  of  clothes,  each  containing  6  yd.  3£  qr., 
can  be  cut  from  333  yards  of  cloth  ?  Ans.    48. 

2.  A  man  bought  a  gold  chain,  weighing  1  oz.  15  pwt.,  at 
seven  dimes  a  pennyweight;  what  did  it  cost?      Ans.  $24.50. 

3.  A  physician,  having  2  Ib  3§  55  19  10  gr.  of  medicine, 
dealt  it  out  in  prescriptions  averaging  15  grains  each  ;  how 
many  prescriptions  did  it  make  ?  Ans.   886. 

4.  A  man  bought  1  T.  11  cwt.  12  Ibs.  of  hay,  at  1£  cents 
a  pound ;  what  did  it  cost  ?  Ans.    $38.90. 

5.  What  will  be  the  cost  of  a  load  of  oats  weighing  1456 
pounds,  at  37£  cents  per  bushel?  Ans.    $17.0625. 

6.  If  one  bushel  of  wheat  will  make  45  pounds  of  flour,  how 
many  barrels  will  1000  bushels  make  ?     Ans.  229-bbl.  116  Ib. 

7.  A  load  of  wheat   weighing  2430  pounds  is  worth  how 
much,  at  $1.20  a  bushel?  Ans.    $48.60. 

8.  Paid  $12.50  for  a  barrel  of  beef;  how  much  was  that 
per  pound  ?  Ans.    6£  cents. 

9.  If  a  silver  dollar  measure  one  inch  in   diameter,  how 
many  dollars,  laid  side  by  side  on  the  equator,  would  reach 
round  the  earth  ?  Am.   1573862400. 

10.  In  10  mi.  7  fur.  4  ch.  70 1.,  how  many  links  ? 

Ans.   87470  links. 


DENOMINATE   FRACTIONS.  183 

11.  What  is  the  value  of  a  city  lot,  25  feet  wide  and  100 
feet  long,  if  every  square  inch  is  worth  one  cent?   Ans.  $3600. 

12.  How  many  cords  of  wood  can  be  piled  in  a  shed  50  ft. 
long,  25  ft.  wide,  and  10  ft.  high  ?     Ans.  97  Cd.  5  cd.  ft.  4  cu.  ft. 

13.  A  cistern  10  feet  square  and  10  feet  deep,  will  hold 
how  many  hogsheads  of  water?     Ans.  118  hhd.  46^  gal. 

14.  A  bin  8  feet  long,  5  feet  wide,  and  4J  feet  high,  will 
hold  how  many  bushels  of  grain  ?  Ans.    144T9j  bu. 

15.  How  many  seconds    less   in  every  Autumn  than   in 
either  Spring  or  Summer?  Ans.   86400  sec. 

16.  If  a  person  could  travel  at  the  rate  of  a  second  of  dis- 
tance in  a  second  of  time,  how  much  time  would  he  require  to 
travel  round  the  earth?  Ans.    15  days. 

17.  How  many  yards  of  carpeting,  1  yd.  wide,  will  be  re- 
quired to  carpet  a  room  20  ft.  long  and  18  ft.  wide  ?    Ans.  40. 

18.  A  printer  calls  for  4  reams  10  quires  and  10  sheets  of 
paper  to  print  a  book ;  how  many  sheets  does  he  call  for  ? 

Ans.   2170. 

19.  How  many  times  will  a  wheel,  16  ft.  6  in.  in  circumfer- 
ence, turn  round  in  running  42  miles  ?  Ans.    13440. 

20.  How  many  days,  working  10  hours  a  day,  will  it  re- 
quire for  a  person  to  count  $10000,  at  the  rate  of  one  cent 
each  second  ?  Ans.    27  da.  7h.  46  min.  40  sec. 

21.  A  town,  6  miles  long  and  4^  miles  wide,  is  equal  to 
how  many  farms  of  80  acres  each  ?  Ans.    216. 

22.  At  $21.75  per  rod,  what  will  be  the  cost  of  grading 
10  mi.  176  rds.  of  road  ?  Ans.  $73428. 


REDUCTION   OF  DENOMINATE  FRACTIONS. 

CASE    I. 

211.     To   reduce    a   denominate    fraction   from  a 
greater  to  a  less  unit. 

1.    Reduce  -fa  of  a  bushel  to  the  fraction  of  a  pint. 
Case  I  is  what  ? 


184  REDUCTION. 

OPERATION. 

J£  x  £  X  f  X  f  =V>  -^5-  ANALYSIS.    To  reduce  bushels 

Q  to  pints,  we  must  multiply  by  4, 

,-          '  8,   and   2,   the    numbers   in   the 

$0    1  scale.     And  since  the  given  num- 

4  ber  is  a  fraction  of  a  bushel,  we 

<g  indicate  the  process  as  in  multi- 

g  plication   of  fractions,  and   after 
canceling,  obtain  -f,  the  Answer.. 

4  =  £  pt.,  Ans.  Hence, 

RULE.  Multiply  the  fraction  of  the  higher  denomination  by 
the  numbers  in  the  scale  successively,  between  the  given  and  the 
required  denominations. 

NOTE.     Cancellation  may  be  applied  wherever  practicable. 
EXAMPLES    FOR    PRACTICE. 

2.  Reduce  T^\7Q-  of  a  £  to  the  fraction  of  a  penny. 

Ans.   •£%  d. 

3.  Reduce  T^^  of  a  week  to  the  fraction  of  a  minute. 

Ans.    -j7^  min. 

4.  What  part  of  a  gill  is  ^W  of  a  hosghead  ?   Ans.  ^  gi. 

5.  What  fraction  of  a  grain  is  ¥£^  of  an  ounce  ?    Ans.  ±  gr. 

6.  Reduce  yuutjuzjiy  °f  a  m^e  t°  the  fraction  of  an  inch. 

Ans.    T^fr^s  in. 

7.  Reduce  f  of  £  of  2  pounds  to  the  fraction  of  an  ounce 
Troy.  Ans.  f  oz. 

8.  Reduce  ^£<y  of  a  hogshead  to  the  fraction  of  a  pint. 

Ans.   ff  pt. 

9.  Reduce  y^^  of  an  acre  to  the  fraction  of  a  rod. 

Ans.  $  rd. 

CASE   II. 

212.  To  reduce  a  denominate  fraction  from  a  less 
to  a  greater  unit. 

1.    Reduce  |  of  a  pint  to  the  fraction  of  a  bushel. 


Give  explanation.     Rule.     Case  II  is  what  ? 


DENOMINATE   FRACTIONS.  185 

OPERATION. 

ANALYSIS.       To      reduce 
_^1_1J  .  pints   to   bushels,   we    must 

5       2~84~80'  ^  divide  by  2'  8'  and  4'  the 

numbers  of  the  scale.     And 


Or,      5 

2 
8 


80 


1  —       bu.,  Am. 


since  the  given  number  of 
pints  is  a  fraction,,  we  indi- 
cate the  process,  as  in  divis- 
ion of  fractions,  and  cancel- 
ing, obtain  -fa,  the  Answer* 


RULE.  Divide  the  fraction  of  the  lower  denomination  by  the 
numbers  in  the  scale,  successively,  between  the  given  and  the 
required  denomination. 

NOTE.     The  operation  will  frequently  be  shortened  by  cancellation. 
EXAMPLES    FOR   PRACTICE. 

2.  What  part  of  a  rod  is  ^  of  a  foot  ?  Ans.   y^  rd. 

3.  What  part  of  a  pound  is  -f  of  a  dram  ?       Ans.   T/F^  lb. 

4.  Reduce  £  of  a  cent  to  the  fraction  of  an  eagle. 

Ans.    Yifa,  E. 

5.  A  hand  is  £  of  a  foot  ;  what  fraction  is  that  of  a  mile? 


6.  Reduce  ^  of  2  pwt.  to  the  fraction  of  a  pound.  Ans.  ^|w  lb. 

7.  How  much  less  is  f  of  a  pint  than  £  of  a  hogshead  ? 

Ans.   tJJhhd. 

8.  In  f  of  an  inch  what  fraction  of  a  mile  ?  Ans.  TTr5^ff  <j-  mi. 

9.  f  of  an  ounce  Troy  is  f  of  what  fraction  of  2  pounds  ? 
10.   f  of  an  ounce  is  £  of  what  fraction  of  2  pounds  Troy? 

CASE   III. 

S13.    To  reduce  a  denominate  fraction  to  integers 
of  lower  denominations. 

1.    What  is  the  value  of  f  of  a  hogshead  of  wine  ? 

Give  explanation.    Rule.     Case  LTI  is  what  ? 


186 


REDUCTION. 


OPERATION. 

f  hhd.  X  63  =  -2|JL  gal.  =  39 J  gal. 

f  gal.  X  4  =  -tf  qt.  =  If  qt ;  f  qt.  X  2  =  f  pt,  ~  1  pt. 

Ans.    39  gal.  1  qt.  1  pint. 

ANALYSIS,  f  hhd.  —  |  of  63  gal.,  or  39|  gal. ;  and  f  gal.  —  f  of 
4  qt.,  or  If  qt.  ;  and  f  qt.  =  •£  of  2  pt.,  or  1  pt.  Hence, 

RULE.  I.  Multiply  the  fraction  by  that  number  in  the  scale, 
which  will  reduce  it  to  the  next  lower  denomination,  and  if  the 
result  be  an  improper  fraction,  reduce  it  to  a  whole  or  mixed 
number. 

II.  Proceed  with    the  fractional  part,  if  any,  as  before, 
until  reduced  to  the  denominations  required. 

III.  The  units  of  the  several  denominations,  arranged  in 
their  order,  will  be  the  required  result. 

EXAMPLES    FOR   PRACTICE. 

2.  Reduce  f-  of  a  month  to  lower  denominations. 

Ans.    17  da.  3  h.  25  min.  42f-  sec. 

3.  What  is  the  value  of  f  of  a  £  ?         Ans.   8s.  6  d.  3f  far. 

4.  What  is  the  value  off  of  a  bushel  ? 

5.  Reduce  f  of  15  cwt.  to  its  equivalent  value. 

Ans.   12  cwt.  85  Ibs.  11  oz.  6f  dr. 

6.  Reduce  f  of  f  of  a  pound  avoirdupois  to  integers. 

Ans.    4oz.  llff  dr. 

7.  What  is  the  value  of  f  of  an  acre  ?       Ans.   3  R.  13£  P. 

8.  Reduce  £f  of  a  day  to  its  value  in  integers. 

Ans.    16  h.  36  min.  55^  sec. 

9.  What  is  the  value  of  f  of  a  pound  Troy  ? 

10.  What  is  the  value  of  £  of  f>i  tons?  Ans.  4  T.  5  cwt.  5  5$  Ib. 

1 1.  What  is  the  value  of  §  of  3§  acres  ?   Ans.  1  A.  1  R.  20  P. 

CASE    IV. 

SI 4.  To  reduce  a  compound  number  to  a  fraction 
of  a  higher  denomination. 

1.    What  part  of  a  week  is  5  da.  14  h.  24  min.  ? 
Give  explanation.    Rule.     Case  IV  is  what  ? 


DENOMINATE   FRACTIONS.  187 

OPERATION.  ANALYSIS.    To  find 

5  da.  14  h.  24  min.  =  8064  min.  what  part  one  compound 

1  wk.  =  10080  min.  number  is    of  another, 

8964   —  4wk      An*  they  must  be  reduced  to 

TOO B a-  —  5  w  the  same  denomination. 

In  5  da.  14  h.  24  min.  there  are  8064  minutes,  and  in  1  week  there 
are  10080  minutes.  Since  1  minute  is  yo^o  °^  a  week»  8064  min- 
utes is  iVoVo  —  I  of  a  week.  Hence, 

RULE.  Reduce  the  given  number  to  its  lowest  denomination 
for  the  numerator,  and  a  unit  of  the  required  denomination 
to  the  same  denomination  for  the  denominator  of  the  required 
fraction. 

NOTE.  If  the  given  number  contain  a  fraction,  the  denominator  of 
this  fraction  must  be  regarded  as  the  lowest  denomination. 

EXAMPLES    FOR    PRACTICE. 

2.   What  part  of  a  mi.  is  6  fur.  26  rd.  3  yd.  2  ft.  ?   Ans.  £  mi. 
3;    What  fraction  of  a  £  is  13  s.  7  d.  3  far.  ? 

4.  Reduce  10  oz.  lOpwt.  10  gr.  to  the  fraction  of  a  pound 
Troy.  Ans.  %$%  Ib. 

5.  Reduce  2  cd.  ft.  8  en.  ft.  to  the  fraction  of  a  cord. 

Ans.    T\  Cd. 

6.  Reduce  1  bbl.  1  gal.  1  qt.  1  pt.  1  gi.  to  the  fraction  of  a 
hogshead.  Ans.  ££J  hhd. 

7.  What  part  of  2  rods  is  4  yards  1£  feet?  Ans.   -/^. 

8.  Reduce  If  pecks  to  the  fraction  of  a  bushel.   Ans.  f  bu. 

9.  What  part  of  9  feet  square  are  9  square  feet  ? 

10.  From  a  piece  of  cloth  containing  8  yd.  3  qr.  a  tailor  cut 
2  yd.  2  qr. ;  what  part  of  the  whole  piece  did  he  take  ?  Ans.  f. 

CASE  v. 

215.  To  reduce  a  denominate  decimal  to  integers 
of  lower  denominations. 

1 .  Reduce  .78125  of  a  pound  Troy  to  integers  of  lower  de- 
nominations. 

Give  explanation.     Rule.     Case  V  is  what  ? 


188  REDUCTION. 

OPERATION.  ANALYSIS.    We   first  multiply 

.78125  Ib.  bv  12  to  reduce  the  given  number 

12  from  pounds  to   ounces,  and  the 

result  is  9  ounces  and  the  decimal 

9.37500  oz.  .375  Of  an  oz.     We  then  multiply 

20  this  decimal  by  20  to  reduce  it  to 

7.50000  pwt.  pennyweights,  and  get  7  pwt.  and 

24  .5  of  a  pwt.     This  last  decimal  we 

multiply   by   24,  to  reduce  it  to 

12.0000  gr.  grains,   and  the  result  is   12  gr. 

9  oz.  7  pwt.  12  gr.,  Ans.         f*™e  the  answer  is  9  oz'  7  ?wt 

RULE.  I.  Multiply  the  given  decimal  by  that  number  in  the 
scale  which  will  reduce  it  to  the  next  lower  denomination,  and 
point  off  as  in  multiplication  of  decimals. 

II.  Proceed  with  the  decimal  part  of  the  product  in  the  same 
manner  until  reduced  to  the  required  denominations.  The  in- 
tegers at  the  left  will  be  the  answer  required. 

EXAMPLES    FOR   PRACTICE. 

2.  What  is  the  value  of  .217°  ?  Ans.    13'  1.2". 

3.  What  is  the  value  of  .659  of  a  week  ? 

Ans.   4  da.  14  h.  42  min.  43.2  sec. 

4.  Reduce  .578125  of  a  bushel  to  integers  of  lower  denom- 
inations. Ans.   2  pk.  2  qt.  1  pt. 

5.  Reduce  .125  bbl.  to  integers  of  lower  denominations. 

Ans.   3  gal.  3  qt.  1  pt.  2  gi. 

6.  What  is  the  value  of  .628125  £  ? 

7.  What  is  the  value  of  .22  of  a  hogshead  of  molasses  ? 

Ans.    13  gal.  3  qts.  3.52  gi. 

8.  What  is  the  value  of  .67  of  a  league  ? 

Ans.    2  mi.  3  rd.  1  yd.  3$  in. 

9.  What  is  the  value  of  .42857  of  a  month  ? 

Ans.    12  da.  20  h.  34  min.  13££  sec. 
10.   What  is  the  value  of  .78875  of  a  long  ton  ? 

Ans.    15  cwt.  3  qr.  2  Ib.  12.8  oz. 

Give  explanation.    Rule. 


DENOMINATE  FRACTIONS.  189 

11.  What  is  the  value  of  5.88125  acres  ?  Ans.  5  A.  3  R.  21  P. 

12.  Reduce  .0055  T.  to  pounds.  Ans.    11  Ib. 

13.  Reduce   .034375.  of  a  bundle  of  paper  to  its  value  in 
lower  denominations.  Ans.    1  quire  9  sheets. 

CASE    VI. 

216.   To  reduce  a  compound  number  to  a  decimal 
of  a  higher  denomination. 

1.    Reduce  3  pk.  2  qt.  to  the  decimal  of  a  bushel. 

OPERATION.  ANALYSIS.     Since  8  quarts  make 

2.00      qt.  1  peck,  and  4  pecks  1  bushel,  there 

~~7~~~  will  be  |  as  many  pecks  as  quarts 

(183),    and  J  as  many  bushels  as 


8 


.8125  bu.,  Ans.     pecks. 

Or  we  may  reduce  3  pk.  2  qt.  to 

Or,  3  pk.  2  qt.  =    56  qt.        the  fraction  Of  a  bushel  (as  in  214), 
1  bu.         =  32  qt.         and  we  aave  |_|  Of  a  bushel,  which, 
ff  =  .8125  bu.,  Ans.     reduced  to  a  decimal,  equals  .8125. 
Hence  the 

RULE.  Divide  the  lowest  denomination  given  by  that  num- 
ber in  the  scale  which  will  reduce  it  to  the  next  higher,  and  an- 
nex the  quotient  as  a  decimal  to  that  higher.  Proceed  in  the 
same  manner  until  the  whole  is  reduced  to  the  denomination 
required.  Or, 

Reduce  the  given  number  to  a  fraction  of  the  required  de- 
nomination, and  reduce  this  fraction  to  a  decimal. 

EXAMPLES    FOR    PRACTICE. 

2.  Reduce  3  qt.  1  pt.  1  gi.  to  the  decimal  of  a  gallon. 

Ans.   .90625  gal. 

3.  Reduce  10  oz.  13  pwt.  9  gr.  to  the  decimal  of  a  pound 
Troy.  Ans.  .8890625  Ib. 

4.  Reduce  1.2  pints  to  the  decimal  of  a  hogshead. 

Ans.  .00238 +  hhd. 

5.  What  part  of  a  bushel  is  3  pk.  1,12  qt.  ?      Ans.  .785  bu. 


Case  VI  is  what  ?     Give  explanations.     Rule. 


190  ADDITION. 

6.  What  part  of  an  acre  is  3  R.  12.56  P.  ? 

7.  Reduce  17  yd.  1  ft.  6  in.  to  the  decimal  of  a  mile. 

Ans.  .00994318  + mi. 

8.  Reduce  .32  of  a  pint  to  the  decimal  of  a  bushel. 

Ans.  .005  bu. 

9.  Reduce  4|  feet  to  the  decimal  of  a  fathom. 

Ans.  .8125  fathom. 

10.  Reduce  150  sheets  of  paper  to  the  decimal  of  a  ream. 

Ans.  .3125  Rm. 

11.  Reduce  47.04  Ib.  of  flour  to  the  decimal  of  a  barrel. 

12.  Reduce  .33  of  a  foot  to  the  decimal  of  a  mile. 

13.  Reduce  5  h.  36  min.  57T6<y  sec.  to  the  decimal  of  a  day. 


ADDITION. 

1.  A  miner  sold  at  one  time  10  Ib.  4oz.  16  pwt.  8  gr. 
of  gold  ;  at  another  time,  2  Ib.  9  oz.  3  pwt. ;  at  another,  1 1  oz. 
20  gr. ;  and  at  another,  25  Ib.  16  pwt.  23  gr. ;  how  much  did 
he  sell  in  all  ? 

OPERATION.  ANALYSIS.    Arranging  the  num- 

ib.       oz.     pwt.      gr.  hers  in  columns,  placing  units  of  the 

10        4     16        8  same  denomination  under  each  oth- 

2930  er>  we   first   add  the  units  in  the 

0     11        0     20  right  hand  column,  or  lowest  de- 

o1- »     f\      IP      OQ          nomination,  and  find  the  amount  to 

zo        u      ID      zo          •      -«          •  ,  .  ,     .  i         n 

be  51   grains,  which  is  equal  to  2 

Ans.  39        1.    17        3          pwt.  3gr.     We  write  the  3  gr.  under 

the  column  of  grains,  and  add  the  2 

pwt.  to  the  column  of  pwt.  We  find  the  amount  of  the  second  col- 
umn to  be  37  pwt.,  which  is  equal  to  1  oz.  17  pwt.  Writing  the  17 
pwt.  under  the  column  of  pwt.,  we  add  the  1  oz.  to  the  next  column. 
Adding  this  column  in  the  same  manner  as  the  preceding  ones,  we 
find  the  amount  to  be  25  oz.,  equal  to  2  Ib.  1  oz.  Placing  the  1  oz. 
under  the  column  of  oz.,  we  add  the  2  Ib.  to  the  column  of  Ib. 
Adding  the  last  column,  we  find  the  amount  to  be  39  Ib.  Hence 
the  following 

What  is  addition  of  compound  numbers  ?     Give  explanation. 


COMPOUND  NUMBERS.  191 

RULE.  I.  Write  the  numbers  so  that  those  of  the  same  unit 
value  will  stand  in  the  same  column. 

II.  Beginning  at  the  right  hand,  add  each  denomination  as 
in  simple  numbers,  carrying  to  each  succeeding  denomination 
one  for  as  many  units  as  it  takes  of  the  denomination  added ',  to 
make  one  of  the  next  higher  denomination. 

EXAMPLES    FOR   PRACTICE. 

(2.)  (3.) 

£.  s.  d.  Ib.  §.  3.  9.  gr. 

48  13  8  12   8  7  2  15 

51  6  4  10  4  1  10 

67  11  3  15  00  2  1  19 

76  18  10  11  6  0  12 

244  10  1  13   4  4  2  00 

(4.)  (5.) 

T.    cwt.    Ib.  oz.     dr.  bu.   pk.  qt.   pt. 

4       7     18  4     10  1371 

15     98  15       5  3220 

3       9     10  6     15  161 

10     15  04  17     051 

9     12     42     11       2  45     2     4     ° 

6.  What  is  the  sum  of  4  mi.  3  fur.  30  rd.  2  yd.  1  ft.  10  in., 
5  mi.  6  fur.  18  rd.  1  yd.  2  ft.  6  in.,  10  mi.  4  fur.  25  rd.  2  yd. 
2  ft.  11  in.,  and  6  fur.  28  rd.  4  yd.  2  ft.  1  in.  ? 

7.  Find  the  sum  of  197  sq.  yd.  4  sq.ft.1  104£  sq.  in.,  122 
sq.  yd.  2  sq.  ft.  27 j  sq.  in.,  5  sq.  yd.  8  sq.  ft.  2|  sq.  in.,  and  237 
sq.  yd.  7  sq.  ft.  128£  sq.  in.  ? 

Ans.  563  sq.  yd.  4  sq.  ft.  118.825  sq.  in. 

NOTE.  When  common  fractions  occur,  they  should  be  reduced  to  a 
common  denominator,  to  decimals,  or  to  integers  of  a  lower  denomi- 
nation, and  added  according  to  the  usual  method. 

Give  the  Rule. 


192  ADDITION. 


A. 
26 
19 
456 

R.       P. 

3      28 
2      38 
2      20 

(8.) 
sq.  yd. 
25 
30 
16 

sq.  ft.     sq.  in. 
8       125 
7       150 
6        98 

503 

1        8 

12Q-)  5        85 
(i)  =  *(*) 

503 

1        8 

13 

1         13 

mi.    fur. 
1       7 

(9.) 
rd.      yd.      ft. 
30        4        2 

in. 
11 

hhd. 
27 

(10.) 

gal.      qt. 
65        3 

pt. 

2 

3 

4 

00 

2       1 

10 

112 

60 

2 

3 

10 

7 

25 

1       2 

11 

50 
421 
14 

29 
00 
39 

0 
2 
1 

1 
3 
2 

16 

bu. 
23 

3       16 

(11.) 

pk.       qt. 
3         7 

4       0 

pt. 
1 

2 

yr. 

25 

(12.) 
da.        h. 
300       19 

min. 
54 

sec. 
35 

34 

2 

0 

1 

21 

40 

12 

40 

24 

42 

3 

5 

0 

3 

112 

14 

15 

17 

51 

1 

4 

1 

6 

19 

11 

45 

59 

23 
11 

0 
3 

3 
4 

0 
0 

1 

1 

1 

1 

1 

57 

109      11      37      16 

bundles  1  ream  15  quires 
bundles  1  ream  10  quires 
13  sheets,  how  much  does 

13.    If  a  printer 
'  sheets  of  paper, 
'   sheets,  and  the 

one  day  use  4 
the  next  day  3 
next  2  bundles 

10 

he  use  in  the  three  days  ? 

Am.    2  bales  1  ream  6  quires  19  sheets. 
14.    A  tailor  used,  in  one  year,  2  gross  5  doz.  10  buttons, 
another  year  3  gross  7  doz.  9,  and  another  year  4  gross  6 
doz.  11  ;  how  many  did  he  use  in  the  three  years? 

Ans.    10  gross  8  doz.  6. 


COMPOUND   NUMBERS.  193 

15.  A  ship,  leaving  New  York,  sailed  east  the  first  day  3° 
45'  50";  the  second  day,  4°  50'  10''  ;  the  third,  2°  10'  55"; 
the  fourth,  2°  39"  ;  how  far  was  she  then  east  from  the  place 
of  starting?  Arts.    12°  47'  34". 

16.  A  man,  in  digging  a  cellar,  removed  127  cu.  yd.  20  cu. 
ft.  of  earth  ;    in  digging  a  drain,  6  cu.  yd.  25  cu.  ft.  ;  and  in 
digging  a  cistern,  17  cu.  yds.  18  cu.  ft.  ;  what  was  the  amount 
of  earth  removed,  and  what  the  cost  at  16  cents  a  cu.  yd.? 

Ans.    152£  cu.  yds.  ;  $24.37£. 

17.  A  farmer  received  80  cents  a  bushel  for  4  loads  of 
corn,  weighing  as  follows  :  2564,  2713,  3000,  and  3109  Ibs.  ; 
how  much  did  he  receive  for  the  whole?  Ans.    $162.657-(- 

18.  A  druggist  sold  for  medicine,  in  three  years,  at  an  aver- 
age price  of  9  cents  a  gill,  the  following  amounts  of  brandy, 
viz.  :   1   bbl.  4  gal.   1  pt.  ;    30  gal.  2  qt,  1  gi.  ;    2  bbl.  15  gal.  ; 
how  much  did  he  receive  for  the  whole?        Ans.    $415.17. 

218.    To  add  denominate  fractions. 

1.  Add  |-  of  a  mile  to  -J  of  a  furlong. 

OPERATION.  ANALYSIS.    We  find  the 

£  mi.  —  6  fur.  26  rd.  11    ft.          value  of  each  fraction  in  in- 
\  f      _  13  rd      5-L  ft  tegers  of  less  denominations 

(213),  and  then  add  their 
Ans.  7  fur.  00  0  values  as  in  compound  num- 

Or,  J  fur.  -j-  8  =  2V  m>-  bers  (  217  }' 

A  mi.  +  1  mi.  =  &  mi.  =  7  fur.      .  °%  we,  .  "»*  r.e<Juce    the, 

given  fractions  to  fractions  of 

the  same  denomination  (212),  then  add  them,  and  find  the  value 
of  their  sum  in  lower  denominations  (213). 

2.  Add  f  of  a  rod  to  J  of  a  foot.  Ans.    13  ft.  1£  in. 

3.  What  is  the  sum  of  £  of  a  mile,  f  of  a  furlong,  and  £  of 
a  rod  ?  Ans.    7  fur.  27  rd.  8  ft.  3  in. 

4.  What  is  the  sum  of  f  of  a  pound  and  f  of  a  shilling  ? 

Ans.    13  s.  10  d.  2f  qr. 

5.  What  is  the  sum  of  |  of  a  ton  and  ^  of  1  cwt.  ? 

Ans.    12  cwt.  42  Ib.  13f  oz. 


Give  explanation  of  the  process  of  adding  denominate  fractions. 
Q 


194  SUBTRACTION. 

6.  What  is  the  sum  of  |-  of  a  day  added  to  ^  an  hour  ? 

Ans.    9  h.  30  min. 

7.  What  is  the  sum  of  £  of  a  week,  f  of  a  day,  and  £  of 
an  hour  ?  Ans.    1  da.  22  h.  15  min. 

8.  Add  f  of  a  hhd.  to  f  of  a  gal. 

9.  What  is  the  sum  of  f-  of  a  cwt.,  8f  lb.,  and  3T9^  oz.  by 
long  ton  table  ?  Ans.    73  lb.  1  oz.  3fi  dr. 

10.  What  is  the  sum  of  f  of  a  mile,  f  of  a  yard,  and  f-  of 
a  foot  ? 

11.  Sold    4  village  lots ;  the  first  contained  £  of  £  of  an 
acre ;  the  second,  60 J  rods ;  the  third,  f  of  an  acre ;  and  the 
fourth,  f  of  §  of  an  acre ;  how  much  land  in  the  four  lots  ? 

Ans.    3  R.  26  P.  126T4T5?  sq.  ft. 

12.  A  farmer  sold  three  loads  of  hay ;  the  first  weighed 
1£  T.,  the  second,  1-^  T.,  and  the  third,  18f  cwt. ;  what  was 
the  aggregate  weight  of  the  three  loads  ? 

Ans.   3  T.  5  cwt.  91  lb.  10§  oz. 

SUBTRACTION. 

219.     1.   If  a  druggist  buy  25  gal.  2  qt.  1  pt.  1  gi.  of 
wine,  and  sell  18  gal.  3qt.  1  pt.  2  gi.,  how  much  has  he  left? 

OPERATION.  ANALYSIS.     Writing  the  subtrahend 

gal.     qt.  .pt.    gi.         under  the  minuend,  placing  units  of  the 

25     2     1      1          same  denomination  under  each  other, ' 

18     3     0     2         we  begin  at  the  right  hand,  or  lowest 

A      ~7>     Q 3~       denomination ;    since  we   cannot   take 

2  gi.  from  1  gi.,  we  add  1  pt.  or  4  gi.  to 

1  gi.,  making  5  gi. ;  and  taking  2  gi.  from  5  gi.,  we  write  the  remain- 
der, 3  gi.,  underneath  the  column  of  gills.  Having  added  1  pt.  or 
4  gi.  to  the  minuend,  we  now  add  1  pt.  to  the  0  pt.  in  the  subtra- 
hend, making  1  pt. ;  and  1  pt.  from  1  pt.  leaves  0  pt.,  which  we  write 
in  the  remainder.  Next,  as  we  cannot  take  3  qt.  from  2  qt.,  we  add 
1  gal.  or  4  qt.  to  2  qt.,  making  6  qt,  and  taking  3  qt.  from  6  qt.,  we 
write  the  remainder,  3  qt.,  under  the  denomination  of  quarts.  Add- 
ing 1  gal.  to  18  gal.,  we  subtract  19  gal.  from  25  gal.,  as  in  simple 

What  is  subtraction  of  compound  numbers  ?    Give  explanation. 


COMPOUND  NUMBERS. 


195 


numbers,  and  write  the  remainder,  6  gal.,  under  the  column  of  gal- 
lons.    Hence  the  following 

RULE.     I.     Write  the  subtrahend  under  the  minuend,  so  that 
units  of  the  same  denomination  shall  stand  under  each  other. 

II.  Beginning  at  the  right  hand,  subtract  each  denomination 
separately,  as  in  simple  numbers. 

III.  If  the  number  of  any  denomination  in  the  subtrahend 
exceed  that  of  the  same  denomination  in  the  minuend,  add  to 
the  number  in  the  minuend  as  many  units  as  make  one  of  the 
next  higher  denomination,  and  then  subtract ;  in  this  case  add 
1   to  the  next  higher  denomination  of  the  subtrahend  before 
subtracting.     Proceed  in  the  same  manner  with  each  denomi- 
nation. 

EXAMPLES    FOR   PRACTICE. 

(2.)  (3.) 

lb.      oz.    pwt.  gr.  A.  B.        P. 

From     18      6      10  14  25  2      16.9 

Take      10      5        4  6  19  3      25.14 


Hem. 

8168 

5      2      31.76 

W 

(5.) 

T. 

cwt. 

lb. 

yr- 

da. 

h. 

min. 

sec. 

14 

11 

69f 

38 

187 

16 

45 

50 

10 

12 

98-| 

17 

190 

20 

50 

40 

20       361       19       55       10 

6.  A  Boston  merchant  bought  English  goods  to  the  amount 
of  4327  £  13s.  7£d.,  and  he  paid  1374  £  10s.  11}  d.;  how 
much  did  he  then  owe  ? 

7.  From  300  miles  take  198  mi.  7  fur.  25  rd.  2yd.   1ft. 
10  in.  Ans.    101  mi.  14  rd.  2  yd.  2  ft.  8  in. 

8.  What  is  the  difference  in  the  longitude  of  two  places, 
one  75°  20'  30"  west,  and  the  other  71°  19'  35"  west? 

Ans.   4°  55". 

9.  From  10  ft  7  §  4  3  1  9  15  gr.  take  3ft8§2329 
18  gr.  Ans.    6  ft  11  §  1    3  1  9  17  gr. 

Give  the  Rule. 


196  SUBTRACTION. 

10.  The  apparent  periodic  revolution  of  the  sun  is  made  in 
365  da.  6  h.  9  min.  9  sec.,  and  that  of  the  moon  in  29  da.  12  h. 
44  mill.  3  sec. ;  what  is  the  difference  ? 

Ans.    335  da.  15  h.  25   min.  6  sec. 

11.  A  man,  having  a  hogshead  of  wine,  drank,  on  an  aver- 
age, for  five  years,  including  two  leap  years,  one  gill  of  wine 
a  day ;  how  much  remained  ?        Ans.  5  gal.  3  qt.  1  pt.  1  gi. 

12.  A  section   of  land  containing  640   acres  is  owned  by 
four  men  ;  the  first  owns  196  A.  2  R.  16£  P. ;  the  second,  200 
A.  1£  R. ;  the  third,  177  A.  36  P. ;  how  much  does  the  fourth 
own  ?  Ans.    65  A.  3  R.  7.75  P. 

13.  From  a  pile  of  wood  containing  75 J    Cd.  was    sold 
at  one  time  16  Cd.  5  cd.  ft.;  at  another,  24  Cd.  6cd.ft.  12 
cu.  ft. ;  at  another,  27  Cd.  112  cu.  ft. ;  how  much  remained  in 
the  pile  ?  Ans.    6  Cd.  3  cd.  ft.  4  cu.  ft. 

14.  If  from  a  hogshead  of  molasses  10  gal.  1  qt.  1  pt,  be 
drawn  at  one  time,  15  gal.  1  pt.  at  another,  and  14  gal.  3  qt. 
at  another,  how  much  will  remain  ? 

22O.    To  find  the  difference  in  dates. 

1.  What  length  of  time  elapsed  from  the  discovery  of 
America  by  Columbus,  Oct.  14,  1492,  to  the  Declaration  of 
Independence,  July  4,  1776  ? 

FIRST  OPERATION.  ANALYSIS.     We  place  the  earlier  date 

yr.  mo.       da.          unde»  the  later,  writing  first  on  the  left 

177b          7  the  number  of  the  year  from  the  Chris- 

1492        10        14          tian  era,  next  the  number  of  the  month, 

2£o  o TT  counting  January  as  the  first  month,  and 

next  the  number  of  the  day  from  the 

first  day  of  the  month.     Instead  of  the  number  of  the  year,  month, 
and  day,  some  use  the  number  of  years,   months,  and  days   that 

SECOND   OPERATION.          JmVe  el°PSed  sinCG  tnC  C<nristian  era» tnus  : 

vr>  *         wt ""      fla*  '        instead  of  saying  July  is  the  7th  month, 

1775          6          3          we    say   ^    months    and    3    days    have 

1491  0        13          elapsed,  and  instead  of  saying  October 

is  the  10th  month,  we  say  9  months  and 

8       20         13  days  have  elapsed. 

How  is  the  difference  of  dates  found  ? 


COMPOUND  NUMBERS. 


197 


Both  methods  will  obtain  the  same  result ;  the  former  is  generally 
used. 

NOTES.  1.  When  hours  are  to  be  obtained,  we  reckon  from  12  at 
night,  and  if  minutes  and  seconds,  we  write  them  still  at  the  right  of 
hours. 

2.  In  finding  the  time  between  two  dates,  or  in  computing  interest, 
12  months  are  considered  a  year,  and  30  days  a  month. 

When  the  exact  number  of  days  is  required  for  any  period 
not  exceeding  one  ordinary  year,  it  may  be  readily  found  by 
the  following 

TABLE, 

Showing  the  number  of  days  from  any  day  of  one  month  to  the  same  day 
of  any  other  month  within  one  year. 


FROM  ANY 
DAY  OP 

TO  THE  SAME  DAY  OF  THE  NEXT. 

Jan. 

Feb. 

Mar. 

Apr. 

May. 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

January  .  .  . 
February  .  . 
March  .... 
April  .... 

365 
334 
306 
275 
245 
214 
'184 
153 
122 
92 
61 
31 

31 
365 
337 
306 
276 
245 
215 
184 
153 
123 
92 
62 

59 
28 
365 
334 
304 
273 
243 
212 
181 
151 
120 
90 

90 
59 
31 
365 
335 
304 
274 
243 
212 
182 
151 
121 

120 
89 
61 
30 
365 
334 
304 
273 
242 
212 
181 
151 

151 

120 
92 
61 
31 
365 
335 
304 
273 
243 
212 
182 

181 
150 
122 
91 
61 
30 
365 
334 
303 
273 
242 
212 

212 
181 
153 
122 
92 
61 
31 
365 
334 
304 
273 
243 

243 
212 
184 
153 
123 
92 
62 
31 
365 
335 
304 
274 

273 
242 
214 
183 
153 
122 
92 
61 
30 
365 
334 
304 

304 
273 
245 
214 
184 
153 
123 
92 
61 
31 
365 
335 

334 
303 
275 
244 
214 
183 
153 
122 
91 
61 
30 
365 

May 

June  .  . 

July 

August  .  .  . 
September  . 
October.  .  .  . 
November  . 
December.  . 

If  the  days  of  the  different  months  are  not  the  same,  the 
number  of  days  of  difference  should  be  added  when  the  earlier 
day  belongs  to  the  month  from  which  we  reckon,  and  subtracted 
when  it  belongs  to  the  month  to  which  we  find  the  time.  If 
the  29th  of  February  is  to  be  included  in  the  time  computed, 
one  day  must  be  added  to  the  result. 

EXAMPLES    FOR    PRACTICE. 

2.  George  Washington  was  born  Feb.  22,  1732,  and  died 
Dec.  14  1799  ;  what  was  his  age  ?  Ans.  67  yr.  9  mo.  22  da. 

How  can  the  number  of  days,  if  less  than  a  year,  be  obtained  ? 
Q* 


198  SUBTRACTION. 

3.  How   much  time  has   elapsed  since  the  declaration  of 
independence  of  the  United  States  ? 

4.  How  many  years,  months,  and  days  from  your  birthday 
to  this  date  ;  or  what  is  your  age  ? 

5.  How  long  from  the  battle  of  Bunker  Hill,  June  17,  1775, 
to  the  battle  of  Waterloo,  June  18,  1815  ?      Am.  40  yr.  1  da. 

6.  What  length  of  time  will  elapse  from  20  minutes  past 
2  o'clock,  P.  M.,  June  24,  1856,  to  10  minutes  before  9  o'clock, 
A.  M.,  January  3,  1861  ?     Ans.  4  yr.  6  mo.  8  da.  18  h.  30  min. 

7.  How  many  days  from  any  day  of  April  to  the  same  day 
of  August?  of  December?  of  February? 

8.  How  many  days  from  the  6th  of  November  to  the  15th 
of  April?  Ans.    160  days. 

9.  How  many  days  from  the  20th  of  August  to  the  15th 
of  the  following  June  ?  Ans.    299  days. 


To  subtract  denominate  fractions. 

1.  From  -f  of  an  oz.  take  %  of  a  pwt. 

OPERATION.  ANALYSIS.     We  per- 

il oz.     =7  pwt.  12  gr.  form  the    same  reduc- 

i  p-wt.  —  21  gr.  tions  as  in  addition  of 

denominate      fractions, 
6  pwt.  15  gr.,  Ans.  (218  ^  and  then  sub- 

'  tract  the  less  value  from 

Or,      fo».X20=r^pwt.  the  greater. 

Y  —  I-  —  */.  pwt.  =  6  pwt.  15  gr. 

2.  What  is  the  difference  between  ^  rod  and  f  of  a  foot  ? 

Ans.   7ft.  6  in. 

3.  From  £  £  take  §  of  f  of  a  shilling. 

4.  From  f  of  a  league  take  ^  of  a  mile. 

Ans.    1  mi.  2  fur.  16rd. 

5.  From  8-ft-  cwt.  take  1  qr.  2f  Ib. 

Ans.    8  cwt,  2  qr.  14  Ib.  5  oz.  15^\  dr. 

6.  From  J-  of  a  week  take  ±  of  a  day. 

Ans.    1  da.  4  h.  48  min. 


Give  explanation  of  the  process  of  subtracting  denominate  fractions. 


COMPOUND   NUMBERS.  199 

7.  Two  persons,  A  and  B,  start  from  two  places  120  miles 
apart,  and  travel  toward  each  other ;  after  A  travels  f ,  and 
B  f ,  of  the  distance,  how  far  are  they  apart  ? 

Ans.    41.  mi.  7  fur.  9  rd.  8  ft.  7£  in. 

8.  From  a  cask  of  brandy  containing  96  gallons,  -£  leaked 
out,  and  f  of  the  remainder  was  sold ;  how  much  still  remained 
in  the  cask  ?  Ans.    25  gal.  2  qt.  3£  gi. 


MULTIPLICATION. 

1.   A  farmer  has  8  fields,  each  containing  4  A.  2  R. 
27  P. ;  how  much  land  in  all  ? 

OPERATION.  ANALYSIS.     In  8  fields  are  8  times  as  much 

A.     R.      P.  land  as  in  1  field.     We  write  the  multiplier 

4     2     27  under  the  lowest  denomination  of  the  mul~ 

8  tiplicand,  and   proceed  thus;    8  times  27  P- 

— -    ™77~          are    216  P.,    equal  to  5  R.    16  P.  ;    and  we 

write  the  16  P.  under  the  number  multiplied. 

Then  8  times  2  R.  are  16  R.,  and  5  R.  added  make  21  R.,  equal  to 

4  A.  1  R. ;  and  we  write  the   1  R.  under  the  number  multiplied. 

Again,  8  times  4  A  are  32  A.,  and  4  A.  added  make  36  A.,  which  we> 

write  under  the  same  denomination  in  the  multiplicand,  and  the 

work  is  done.     Hence, 

RULE.  I.  Write  the  multiplier  under  the  lowest  denomina- 
tion of  the  multiplicand. 

II.  Multiply  as  in  simple  numbers,  and  carry  as  in  addi- 
tion of  compound  numbers. 

EXAMPLES   FOR    PRACTICE. 


(2.) 

bu.    pk.    qt.    pt. 

4251 

(3.) 

mi.    fur.     rd.        ft. 

9     4    20     13 

2 

6 

9130 

57     3       4     12 

Multiplication  of  compound  numbers,  how  performed  ?    Rule. 


200  MULTIPLICATION. 

(4.)  (5.) 

£.        s.        d.  Ib.    oz.    pwt.    gr. 

5     18     4  3     4     0     22 

4  7 


(6.)  (7.) 

T.        cwt.      Ib.       oz. 

14     16     48     12  13°     10'     35" 

11  9 


8.  In  6  barrels  of  grain,  each  containing  2  bu.  3  pk.  5  qt., 
how  many  bushels  ?  Am.    17  bu.  1  pk.  6  qt. 

9.  If  a  druggist  deal  out  3  Ib  4  §  1  3  2  9  16  gr.  of  med- 
icine a  day,  how  much  will  he  deal  out  in  6  days  ? 

10.  If  a  man  travel  29  mi.    3  fur.  30  rd.  loft,  in  1   day, 
how  far  will  he  travel  in  8  days  ? 

11.  If  a  woodchopper  can  cut  3  Cd.  48  cu.  ft,  of  wood  in  1 
day,  how  many  cords  can  he  cut  in  12  days  ?    Ans.  404  Cd. 

12.  What  is  the  weight  of  48  loads  of  hay,  each  weighing 
1  T.  3  cwt.  50  Ib.  ? 

OPERATION.  ANALYSIS.     When  the  multi- 

T-   cwt-   lb-  plier  is  large,  and  a  composite 

1     3     50  number,  we  may  multiply  by  one 

6  of  the  factors,  and  that  product 

"^      \      ™  by  the  other.     Multiplying  the 

/       1       00    weight  of  6  loads.  .   •,  ,     ,,  ..   ,       ,  , 

weight  of  1  load  by  6,  we  obtain 

the  weight  of  6  loads,  and  the 


56     8     00  weight  of  48  loads.         weight  of  6  loads  multiplied  by 

8,  gives  the  weight  of  48  loads. 

13.  If  1  acre  of  land  produce  45  bu.  3  pk.  6  qt.  1  pt.  of 
corn,  how  much  will  64  acres  produce  ?  Ans.    2941  bu. 

14.  How  much  will  120  yards  of  cloth  cost,  at  1  £  9  s.  8£  d. 
per  yard  ? 

15.  If  $80  will  buy  4  A.  3  R.  26  P.  20  sq.  yd.  3  sq.  ft.  of 
land,  how  much  will  $4800  buy  ?      Ans.    295  A.  10  sq.yd. 

16.  If  a  load  of  coal  by  the  long  ton  weigh  1  T.  6  cwt.  2  qr. 
26  lb.  10  oz.,  what  will  be  the  weight  of  73  loads  ? 

Ans.   97  T.  11  cwt.  3  qr.  11  lb.  10  oz. 


COMPOUND   NUMBERS.  201 

17.  The  sun,  on  an  average,  changes  his  longitude  59'  8.33" 
per  day  ;  how  much  will  be  the  change  in  365  days  ? 

18.  If  1  pt.  3  gi.  of  wine  fill  1  bottle,  how  much  will  be  re- 
quired to  fill  a  great  gross  of  bottles  of  the  same  capacity  ? 

DIVISION, 

223.     1.   If  4  acres  of  land  produce  102  bu.  3  pk.  2  qt.  of 

wheat,  how  much  will  1  acre  produce  ? 

OPERATION.  ANALYSIS.     One  acre  will  produce  ^ 

pt.     bu.     Pk.   qt.   pts.        as  much  as  4  acres.     Writing  the  divi- 

4  )  102     3     2  _       sor  on  the  left  of  the  dividend,  we  divide 

25261          ^  ku.  by  4,  and  we  obtain  a  quotient  of 

25  bu.,  and  a  remainder  of  2  bu.    We 

write  the  25  bu.  under  the  denomination  of  bushels,  and  reduce  the 
2  bu.  to  pecks,  making  8  pk.,  and  the  3  pk.  of  the  dividend  added 
makes  11  pk.  Dividing  11  pk.  by  4,  we  obtain  a  quotient  of  2  pk. 
and  a  remainder  of  3  pk.  ;  writing  the  2  pk.  under  the  order  of 
pecks,  we  next  reduce  3  pk.  to  quarts,  adding  the  2  qt.  of  the 
dividend,  making  26  qt.,  which  divided  by  4  gives  a  quotient  of  6  qt. 
and  a  remainder  of  2  qt.  Writing  the  6  qt.  under  the  order  of 
quarts,  and  reducing  the  remainder,  2  qt.,  to  pints,  we  have  4  pt., 
which  divided  by  4  gives  a  quotient  of  1  pt,  which  we  write  under 
the  order  of  pints,  and  the  work  is  done. 

2.   A  farmer  put  132  bu.  OPERATION. 

1  pk.  of  apples  into  46  barrels  ;  bn-     Pk- 

how  many  bu.  did  he  put  into     46  )  132 
a  barrel?  _?±_ 

40 

4 

When  the  divisor  is  large,  and 
not  a  composite  number,  we  di-  (     P 

vide  by  long  division,  as  shown 
in  the  operation.     From  these  23 

examples  we  derive  the 


184(4qt. 

1^1  Am.  2bu.3pk.4qt. 


Explain  the  process  of  dividing  compound  numbers. 


202  DIVISION. 

RULE.  I.  Divide  the  highest  denomination  as  in  simple 
numbers,  and  each  succeeding  denomination  in  the  same  man- 
ner, if  there  be  no  remainder. 

II.  If  there  be  a  remainder  after  dividing  any  denomina- 
tion, reduce  it  to  the  next  lower  denomination,  adding  in  the 
given  number  of  that  denomination,  if  any,  and  divide  as  be- 
fore. 

III.  The  several  partial  quotients  will  be  the  quotient  re- 
quired. 

NOTES.  1.  When  the  divisor  is  large  and  is  a  composite  number, 
we  may  shorten  the  work  by  dividing  by  the  factors. 

2.  When  the  divisor  and  dividend  are  both  compound  numbers,  they 
must  both  be  reduced  to  the  same  denomination  before  dividing,  and 
then  the  process  is  the  same  as  in  simple  numbers. 

EXAMPLES    FOR    PRACTICE. 

(3.)  (4.) 

£.          s.         d.  T.  cwt.         Ib. 

5  )  25       8       4  7  )  45        15       25 


518  6         10      75 

(5.)  (6.) 

wk.      da.        h.        min. 

4  )3      5      22  _  00  10  )  25°      42'      40" 

~~6      17      30  2       34       16 

7.  Bought  6  large  silver  spoons,  which  weighed  11  oz.  3  pwt.  ; 
what  was  the  weight  of  each  spoon  ? 

8.  A  man  traveled  by  railroad  1  000  miles  in  one  day  ; 
what  was  the  average  rate  per  hour  ? 

Ans.    41  mi.  5  fur.  13  rd.  5  ft.  6  in. 

9.  If  a  family  use  10  bbl.  of  flour  in  a  year,  what  is  the 
average  amount  each  day  ?  Ans.    5  Ib.  5  oz.  144&  dr. 

10.  The  aggregate  weight  of  123  hogsheads  of  sugar  is 
57  T.  19  cwt.  42  Ib.  14  oz.  ;    what  is  the  average  weight  per 
hogshead  ?  Ans.    9  cwt.  42  Ib.  10  oz. 

11.  How  many  times  are  5  £  10  s.  10  d.  contained  in  537  £ 
10s.  10d.?  _  Ans.    97. 

Give  the  rule.  When  the  divisor  is  a  composite  number,  how  may 
•\ve  proceed?  When  the  divisor  and  dividend  are"  both  compound 
numbers,  how  proceed  ? 


COMPOUND  NUMBERS.  203 

12.  A  cellar  50  ft.  long,  30  ft.  wide,  and  6  ft.  deep  was  ex- 
cavated by  5  men  in  6  days ;  how  many  cubic  yards  did  each 
man  excavate  daily?  Arts.    11  cu.  yd.  3  cu.ft. 

13.  If  a  town  5  miles  square  be  divided  equally  into  150 
farms,  what  will  be  the  size  of  each  farm  ? 

Ans.    106  A.  2  R.  26  P.  20  sq.  yd.  1  sq.ft.  72  sq.  in. 

14.  How  many  times  are  4  bu.  3  pk.  2  qt.  contained  in 
336  bu.  3pk.  4qt.?  Ans.    70. 

15.  A  merchant  tailor  bought  4  pieces  of  cloth,  each  con- 
taining 60  yd.  2.25  qr. ;  after  selling  £  of  the  whole,  he  made 
up  the  remainder  into  suits  containing  9  yd.  2  qr.  each ;  how 
many  suits  did  he  make  ?  Ans.   17. 


LONGITUDE  AND  TIME.  c 

SS54.  Every  circle  is  supposed  to  be  divided  into  360 
equal  parts,  called  degrees* 

Since  the  sun  appears  to  pass  from  east  to  west  round  the 
earth,  or  through  360°,  once  in  every  24  hours,  it  will  pass 
through  ^  of  360°,  or  15°  of  the  distance,  in  1  hour ;  and  1°  of 
distance  in  y1^  of  1  hour,  or  4  minutes ;  and  V  of  distance  in 
•fa  of  4  minutes,  or  4  seconds. 

TABLE  OF  LONGITUDE  AND  TIME. 

360°  of  longitude  =  24  hours,  or  1  day  of  time. 

15°   "          "          =1  hour  "      « 

1°   "          "         =    4  minutes  "      " 

1'    "          "         =4  seconds  "      " 

CASE    I. 

225.  To  find  the  difference  of  time  between  two 
places,  when  their  longitudes  are  given. 

1.  The  longitude  of  Boston  is  71°  3',  and  of  Chicago  87° 
30; ;  what  is  the  difference  of  time  between  these  two  places  ? 

Explain  how  distance  is  measured  by  time.  Repeat  the  table  of 
longitude  and  time.  Case  I  is  what  r 


204  LONGITUDE   AND   TIME. 

OPERATION.  ANALYSIS.    By  subtraction  of 

87°  30'  compound  numbers  we  first  find 

yj  g/  the  difference  of  longitude  be- 

tween the  two  places,  which  is 

16°          27'  16o  27/.     Since  1°  of  longitude 

'  4  makes  a  difference  of  4  minutes 

,  ,      Z      '.       7Z"  .,  of  time,  and  1'  of  longitude  a 

1  h.  o  ram.  48  sec.,  Ans.  ,._  ,  °  „   . 

difference  of  4  seconds  of  time, 

we  multiply  16°  27'.  the  difference  in  longitude,  by  4,  and  we  obtain 
the  difference  of  time  in  minutes  and  seconds,  which,  reduced  to 
higher  denominations,  gives  1  h.  5  min.  48  sec.,  the  difference  in 
time.  Hence  the 

HULE.  Multiply  the  difference  of  longitude  in  degrees  and 
minutes  by  4,  and  the  product  will  be  the  difference  of  time  in 
minutes  and  seconds,  which  may  be  reduced  to  hours. 

NOTE.  If  one  place  be  in  east,  and  the  other  in  west  longitude,  the 
difference  of  longitude  is  found  by  adding  them,  and  if  the  sum  be 
greater  than  180°,  it  must  be  subtracted  from  360°. 

EXAMPLES    FOR    PRACTICE. 

2.  New  York  is  74°  1'  and  Cincinnati  84°  24'  west  longi- 
tude ;  what  is  the  difference  of  time  ?    Ans.  41  min.  32  sec. 

3.  The  Cape  of  Good  Hope  is  18"  28'  east,  and  the  Sand- 
wich Islands  155J  west  longitude;   what  is  the  difference  of 
time  ?  Ans.    11  h.  33  min.  52  sec. 

4.  Washington   is  77°    V    west,  and   St.  Petersburg  30° 
19'  east  longitude ;  what  is  their  difference  of  time  ? 

Ans.   7  h.  8  min.  20  sec. 

5.  If  Pekin  is   118°  east,  and   San   Francisco  122°  west 
longitude,  what  is  their  difference  of  time  ? 

6.  If  a  message  be  sent  by  telegraph  without  any  loss  of 
time,  at   12  M.  from  London,  0°  0'  longitude,  to  Washington, 
77°  1'  west,  what  is  the  time  of  its  receipt  at  Washington  ? 

NOTE.  Since  the  sun  appears  to  move  from  east  to  west,  when  it  is 
exactly  12  o'clock  at  one  place,  it  will  be  past  12  o'clock  at  all  places 
east,  and  before  12  at  all  places  west.  Hence,  knowing  the  difference 
of  time  between  two  places,  and  the  exact  time  at  one  of  them,  the 
exact  time  at  the  other  will  be  found  by  adding  their  difference  to  Jhe 
given  time,  if  it  be  east,  and  by  subtracting  if  it  be  west. 

Ans.    6  h.  51  min.  56  sec.,  A.  M. 
Give  explanation.    Rule. 


COMPOUND  NUMBERS 


7.  A  steamer  arrives  at  Halifax,  63°  3 1 
P.   M. ;    the  fact  is  telegraphed  to   St. 

without  loss  of  time ;    what  is  the  time  of  itF^^eipt  at  St. 
Louis  ?  Ans.    2  h.  13  min.  24  sec.,  P.  M. 

8.  If,  at  a  presidential  election,  the  voting  begin  at  sunrise 
and  end  at  sunset,  how  much  sooner  will  the  polls  open  and 
close  at  Eastport,  Me.,  67°  west,  than  at  Astoria,  Oregon,  124° 
west  ?  Ans.    3  h.  48  min. 

9.  When  it  was  1  o'clock,  A.  M.,  on  the  first  day  of  Jan- 
uary, 1859,  at    Bangor,    Me.,    68°    47'  west,    what  was  the 
time  at  the  city  of  Mexico,  99°  5'  west? 

Ans.   Dec.  31,  1858,  58  min.  48  sec.  past  10,  P.  M. 

CASE    II. 

226.  To  find  the  difference  of  longitude  between 
two  places,  when  the  difference  of  time  is  known. 

1.  If  the  difference  of  time  between  New  York  and  Cincin- 
nati be  41  min.  32  sec.,  what  is  the  difference  of  longitude  ? 

OPERATION.  ANALYSIS.     Since  4  minutes  of  time 

min.      sec.  make  a  difference  of  1°  of  longitude,  and 

4  )  41       32  4  seconds  of  time,  a  difference  of  I'  of 

~~     ~~     .  longitude,  there  will  be  1  as  many  de- 

'      7  s'  grees  of  longitude  as  there  are  minutes 

of  time,  and  \  as  many  minutes  of  longitude  as  there  are  seconds  of 

time.     Hence, 

RULE.  Reduce  the  difference  of  time  to  minutes  and  sec- 
onds, and  then  divide  by  4 ;  the  quotient  will  be  the  difference 
in  longitude,  in  degrees  and  minutes. 

2.  What  is  the  difference  of  longitude  between  the  Cape 
of  Good  Hope  and  the  Sandwich  Islands,  if  the  difference  of 
time  be  11  h.  33  min.  52  sec.  ?  Ans.    173°  28'. 

3.  What  is  the  difference  of  longitude  between  Washington 
and  St.  Petersburg,  if  their  difference  of  time  be  7  h.  8  min. 
20  sec.?  Ans.    107°  20'. 

Case  II  is  what  ?     Give  explanation.     Rule. 
R 


206  DUODECIMALS. 

4.  When  it  is  half  past  4,  P.  M.,  at  St.  Petersburg,  30°  19' 
east,  it  is  32  min.  36  sec.  past  8,  A.  M.,  at  New  Orleans,  west; 
what  is  the  difference  of  longitude  ?  Am.    119J  21' 

5.  The  longitude  of  New  York  is  74°  1'  west.     A  sea  cap- 
tain leaving  that  port  for  Canton,  with  New  York  time,  finds 
that  his  chronometer  constantly  loses  time.     What  is  his  longi- 
tude when  it  has  lost  4  hours  ?  8  h.  40  min.  ?  13  h.  25  min.  ? 

Ans.    14°  1'  west;  55°  59'  east;  127°  14'  east. 

6.  When  the  days  are  of  equal  length,  and  it  is  noon  on 
the  1st  meridian,  on  what  meridian  is  it  then  sunrise  ?  sun- 
set ?  midnight  ?   Ans.  90°  west ;  90°  east ;  180"  east  or  west. 


DUODECIMALS. 

Duodecimals  are  the  divisions  and  subdivisions  of 
a  unit,  resulting  from  continually  dividing  by  12,  as  1,  y1^,  T£¥, 
TrVs?  &c'  ^n  practice,  duodecimals  are  applied  to  the  meas- 
urement of  extension,  the  foot  being  taken  as  the  unit. 

If  the  foot  be  divided  into  12  equal  parts,  the  parts  are 
called  inches,  or  primes  ;  the  inches  divided  by  12  give  sec- 
onds; the  seconds  divided  by  12  give  thirds;  the  thirds  di- 
vided by  12  give  fourths  ;  and  so  on. 

From  these  divisions  of  a  foot  it  follows  that 

I7     (inch  or  prime)  .........  js    TL.    of  a  foot. 

I"    (second)  or  T^  of  ^,  .....  «  ^  Of  a  foot. 

V"  (third)  or  TV  of  TV  of  TV,  .  .  «  T7VF  of  a  foot,  &c. 


TABLE. 

12  fourths,  marked  (""),  make  1  third  ........  marked  V" 

12  thirds                                «      i  second,  "       1" 

12  seconds                            «     i  prime,  or  inch,  "       V 

12  primes,  or  inches,            "     1  foot,  «       ft. 
SCALE  —  uniformly  12. 

The  marks  ',  ",  '",  "",  are  called  indices. 

What  are  duodecimals  ?    To  what  applied  ?    Explain  the  divisions 
of  the  foot.     Repeat  the  table. 


COMPOUND   NUMBERS.  207 

NOTE.  Duodecimals  are  really  common  fractions,  and  can  always 
be  treated  as  such  ;  but  usually  their  denominators  are  not  expressed, 
and  they  are  treated  as  compound  numbers. 

ADDITION  AND  SUBTRACTION  OF  DUODECIMALS. 

328.  We  add  and  subtract  duodecimals  the  same  as  other 
compound  numbers. 

EXAMPLES. 

1.  Add  13  ft.  4'  8",  10  ft.  6'  7",  145  ft.  9'  II7'.      - 

Ans.    169ft.  9'  2". 

2.  Add  179ft.  11'  4",  245ft.  1'  4",  3ft.  9'  9". 

Ans.   428ft.  10'  5". 

3.  From  25ft.  6'  3"  take  14ft.  9'  8".    Ans.  10ft.  8'  1". 

4.  From  a  board  15  ft.  7'  6"  in  length,  3  ft.  8'  11"  were 
sawed  off;  what  was  the  length  of  the  piece  left? 

Ans.  11  ft.  10'  7". 

MULTIPLICATION  OF  DUODECIMALS. 

229*  Length  multiplied  by  breadth  gives  surface,  and 
surface  multiplied  by  thickness  gives  solid  contents  (108). 

1.  How  many  square  feet  in  a  board  11  feet  8  inches  long 
and  2  feet  7  inches  wide  ? 

OPERATION.  ANALYSIS.    We  first  multiply  by  the  7'. 

lift.     8'  7  twelfths  times  8  twelfths  equals  56  one 

2          7'  hundred     forty-fourths,     which    equals    4 

twelfths  and  8  one  hundred  forty-fourths. 

6  ft-  We  write  the  8  144ths  —  marked  with  two 

23          4'  indices  —  to  the  rigRt,  and  add  the  4  12ths 

30ft       1'     8"  ^°  ^e  nex^  Pr°duct.      7'  times  11   equals 

77',  which  added  to  4'  equals  81',  equal  to 

6  feet  and  9'.    We  write  the  9'  under  the 

inches,  or  12ths,  and  the  6  under  the  feet,  or  units.  2  times  8' 
equals  16',  or  1  foot  and  4'.  We  write  the  4'  under  the  9',  and 
add  the  1  foot  to  the  next  product.  2  times  1 1  feet  are  22  feet,  and 
1  foot  added  make  23  feet,  which  we  write  under  the  6  feet.  Add- 


How  are  duodecimals  added  and  subtracted  ?     Give  analysis  of  ex- 
ample 1. 


208  DUODECIMALS. 

ing  these  partial  products,  and  we  have  30  ft.  V  and  8"  for  the 
entire  product. 

It  will  be  seen  from  the  above  that  the  number  of  indices  to  every 
product  of  any  two  factors  is  equal  to  the  sum  of  the  indices  of  those 
factors  ;  thus  7'  X  &  z=  56"  ;  4'  X  5"'  =  20""'.  Hence  the 

KULE.  I.  Write  the  several  terms  of  the  multiplier  under 
the  corresponding  terms  of  the  multiplicand. 

II.  Multiply  each  term  of  the  multiplicand  by  each  term  of 
the  multiplier,  beginning  with  the  lowest  term  in  each,  and  call 
the  product  of  any  two  denominations  the  denomination  denoted 
by  the  sum  of  their  indices,  carrying  I  for  every  12. 

III.  Add  the  partial  products,    carrying   1  for  every  12; 
their  sum  will  be  the  required  answer. 

EXAMPLES    FOR   PRACTICE. 

2.  How  many  square  feet  in  a  board  13ft.  9'  long  and  11' 
wide?  Ans.    12ft.  7'  3". 

3.  How  many  square  feet  in  a  stock  of  4  boards,  each  11  ft. 
9'  long  and  1  ft.  3'  wide  ?  Ans.   58  ft.  9'. 

4.  How  many  square  yards  of  plastering  on  the  walls  of  a 
room  12ft.  11'  square,  and  9ft.  3'  high,  allowing  for  two  win- 
dows and  one  door,  each  6  ft.  2'  high  and  2  ft.  4'  wide  ? 

Ans.   48  sq.  yd.  2  ft.  9'. 

5.  How  many  solid  feet  in  a  mow  of  hay  30  ft.  4'  long, 
25ft.  6'  wide,  and  12ft.  5'  high?  Ans.    9604ft.  3'  6". 

6.  How  many  cords  in  a  pile  of  wood  18ft.  6'  long,  12ft. 
wide,  and  5  ft.  6'  high  ?  Ans.    9  cords  69  ft. 

7.  How  many  cubic  yards  of  earth  must  be  removed  in 
digging  a  cellar  36  ft.  10'  long,  22  ft.  3'  wide,  and  5  ft.  2'  deep  ? 

Ans.    156cu.yd.  22ft.  3'  1" . 

8.  What  would  it  cost  to  plaster  a  wall  32  ft,  8'  long  and 
9  ft.  high,  at  17  cents  per  square  yard  ?  Ans.    $5.55  £. 

9.  How  many    yards  of   carpeting,  27'  wide,  will  be  re- 
quired to  cover  a  floor  48  ft.  long  and  33  ft.  9'  wide  ? 

Ans.    240  yards. 

Give  the  rule. 


COMPOUND   NUMBERS.  209 

DIVISION  OF  DUODECIMALS. 

23 O.    1.    A  flagstone,  3  ft,  9'  wide,  has  a  surface  of  20  ft. 
II7  3"  ;  what  is  its  length  ? 

OPERATION.  ANALYSIS.    We  divide 

3ft.  9'  )  20ft.  11'     3"  (  5  ft.  7'.         the  surface   by  the  width 

18         9'  to  obtain  the  length.     The 

divisor  is  something  more 

than   3  ft.,  and  to  obtain 

2         2'     3  the  first  quotient  figure,  we 

consider  how  many  times 

3ft.  and  something  more  is  contained  in  nearly  21ft.  (20ft.  11'); 
we  estimate  it  to  be  o  times,  and  multiplying  the  divisor  by  this 
quotient  figure,  we  have  18ft.  9',  which,  subtracted  from  20ft.  11', 
leaves  2  ft.  2',  to  which  we  bring  down  3",  the  last  term  of  the  divi- 
dend. We  next  seek  how  many  times  the  divisor  is  contained  in 
this  remainder,  and  find  by  trial  the  quotient  7' ;  multiplying  the 
divisor  by  this  figure,  we  obtain  2  ft.  2'  3",  and  there  is  no  remain- 
der. Hence  the 

RULE.     I.    Write  the  divisor  on  the  left  hand  of  the  dividend, 
as  in  simple  numbers. 

II.  Find  the  first  term  of  the  quotient  either  by  dividing  the 
first  term  of  the  dividend  by  the  first  term  of  the  divisor,  or  by 
dividing  the  first  two  terms  of  the  dividend  by  the  first  two 
terms  of  the  divisor  ;  multiply  the  divisor  by  this  term  of  the 
quotient,  subtract  the  product  from  the  corresponding  terms  of 
the  dividend,  and  to  the  remainder  bring  down  another  term  of 
the  dividend. 

III.  Proceed  in  like  manner  till  there  is  no  remainder,  or 
till  a  quotient  has  been  obtained  sufficiently  exact. 

EXAMPLES    FOR    PRACTICE. 

2.  Divide  44  ft.  5'  4"  by  16ft.  8'.  Ans.   2  ft.  8'. 

3.  The  square   contents  of  a  walk  are  184  ft.  3',  and  the 
length  is  40  ft.  11'  4"  ;  what  is  the  width?       Ans.    4ft.  6'. 

4.  A  blanket  whose  square  contents  are  14  ft.  6',  is  to  be 
lined  with  cloth  2  ft.  7'  wide ;  how  much  in  length  will  be  re- 
quired ? 

Give  analysis  of  example  1.     Rule. 


210  PROMISCUOUS  EXAMPLES. 

5.  A  block  of  granite  contains  64  ft.  2'  5" ;  its  width  is 
2  ft.  6',  and  its  thickness  3  ft.  7' ;  what  is  its  length  ? 

NOTE.  Since  the  solid  contents  are  the  product  of  the  three  dimen- 
sions, we  divide  the  solid  contents  by  any  two  dimensions  or  by  their 
product,  to  obtain  the  other  dimension. 

Ans.    7  ft.  2'. 

PROMISCUOUS    EXAMPLES. 

1.  In  115200  grains  Troy,  how  many  pounds? 

2.  In  365  da.  5  h.  48  niin.  46  sec.,  how  many  seconds  ? 

Ans.   31556928. 

3.  A  man  wishes  to  ship  1560  bushels  of  potatoes  in  bar- 
rels containing  3  bu.  1  pk.  each ;  how  many  barrels  will  be 
required  ?  Ans,   480. 

4.  Reduce  295218  inches  to  miles. 

5.  Reduce  456575  grains  to  pounds,  apothecaries'  weight. 

Ans.    79  ib  3§  15  IB  15  gr. 

6.  How  many  sheets  in  3  reams  of  paper  ? 

7.  What  is  the  value  of  4  piles  of  wood,  each  20  ft.  long,  6  ft. 
wide,  and  10  ft.  high,  at  $3.25  per  cord  ?      Ans.    $121.87£. 

8.  How  many  bottles,  each  holding  1  qt.  1  gi.,  can  be  filled 
from  a  barrel  of  cider?  Ans.    112. 

9.  At  $26.40  per  sq.  rd.  for  land,  what  will  be  the  cost  of  a 
village  lot  8£  rd.  long,  and  4£  rd.  wide  ?         Ans.    $980.10. 

10.  Divide  259  A.  1  R.  10  P.  of  land  into  36  equal  lots. 

Ans.   7  A.  32J-  P. 

1 1 .  How  many  times  can  a  box  holding  4  bu.  3  pk.  2  qt.  be 
filled  from  336  bu.  3  pk.  4qt.?  Ans.   70. 

1 2.  What  is  the  value  of  .875  of  a  gallon  ? 

13.  What  part  of  a  mile  is  2  fur.  36  rd.  2  yd.  ?   Ans.  •&. 

14.  What  part  of  2  days  is  13  h.  26  min.  24  sec.  ? 

15.  From  26  A.  2  R.  of  land,  5  A.  3  R.  were  sold  ;  what 
part  of  the  whole  piece  remained  unsold  ?  Ans.   -f^. 

]  0.  What  is  the  difference  between  f  of  a  pound  sterling 
and  5£  pence?  Ans.  11  s.  6 £  d. 

1 7.  What  is  the  sum  of  |  of  a  yard,  |  of  a  foot,  and  |  of 
an  inch  ?  Ans.  7  inches. 


PROMISCUOUS  EXAMPLES.  211 

18.  Reduce  3  cwt.  Iqr.  7  Ib.  of  coal  to  the  decimal  of  a 
long  ton.  Ans.    .165625. 

19.  Benjamin  Franklin  was  born  Jan.  18, 1706,  and  George 
Washington   Feb.   22,  1732  j   how  much  older  was  Franklin 
than  Washington  ?  Ans.    26  yr.  1  mo.  4  da. 

20.  The  longitude  of  Boston  is  71°  4'  west,  and  that  of 
Chicago  87°  30'  west;  when  it  is  12  M.  at  Boston,  what  is  the 
time  in  Chicago  ?  Ans.    10  h.  58  min.  16  sec.  A.  M. 

21.  If  the  difference  of  time  between  New  York  and  New 
Orleans  be  1  h.  4  sec.,  what  is  the  difference  in  longitude  ? 

Ans.    15°  1'. 

22.  Add  §  of  a  mile,  £  of  a  furlong,  and   r3?  of  a  rod  to- 
gether. Ans.   5  fur.  33  rd.  8  ft.  3  in. 

23.  If  a  bushel  of  barley  cost  $.80,  what  will  20  bu.  3  pk. 
6  qt.  cost?  Ans.   $16.75. 

24.  What  is  the  value  of  .875  of  a  gross  ?    Ans.    10£  doz. 

25.  How  many  acres  hi  a  field  56^-  rods  long,  and  24.6 
rods  wide  ?  Ans.   8  A.  2  R.  29.9  P. 

26.  How  many  perches  of  masonry  in  the  wall  of  a  cellar 
which  is  20  feet  square  on  the  inside,  8  feet  high,  and  1£  feet 
in  thickness  ?  Ans.    44.6+. 

27.  A,  B.  and  C  rent  a  farm,  and  agree  to  work  it  upon 
shares ;  they  raise  640  bu.  3  pk.  of  grain,  which  they  divide 
as  follows  :  one  fourth  is  given  for  the  rent ;  of  the  remainder 
A  takes  10£  bu.  more  than  one  third,  after  which  B  takes  one 
half  of  the  remainder  less  7  bushels,  and  C  has  what  is  left ; 
how  much  is  C's  share  ?  Ans.    161  bu.  3  pk.  6  qt. 

28.  What  is  the  value  in  Troy  weight  of  13  Ib.  8  oz.  11.4  dr. 
avoirdupois  weight  ?      Ans,  16  Ib.  5  oz.  10  pwt.  11.7  -f-  gr- 

29. .  If  154  bu.  1  pk.  6  qt.  cost  $173.74,  how  much  will  1.5 
bushels  cost?  Ans.   $1.687+. 

30.  What  is  the  value  of  .0125  of  a  ton  ?     Ans.    25  Ibs. 

31.  What  fraction  of  3  bushels  is  TV  of  2bu.  3  pk.  ? 

Ans.    T7/^' 

32.  How  many  wine  gallons  in  a  water  tank  4  feet  long, 
3£  feet  wide,  and  1  ft.  8  in.  deep  ?  Ans.    174^. 


212  PROMISCUOUS   EXAMPLES. 

33.  How  many  bushels  will  a  bin  contain  that  is  7£  feet 
square,  and  6  ft,  8  in.  deep  ?  Ans.    301.336  +  bu. 

34.  How  much,  must  be  paid  for  lathing  and  plastering 
overhead  a  room  36  feet  long  and  20  feet  wide,  at  26  cents  a 
square  yard  ? 

35.  How  many  shingles  will  it  take  to  cover  the  roof  of  a 
building  46  feet  long,  each  of  the  two  sides  of  the  roof  being 
20  feet  wide,  allowing  each  shingle  to  be  4  inches  wide,  and 
to  lie  5  inches  to  the  weather  ?  Ans.    13248. 

36.  John  Young  was  born  at  a  quarter  before  4  o'clock,  A. 
M.,  Sept.  4,  1836;  what  will  be  his  age  at  half  past  6  o'clock, 
P.  M.,  April  20, 1864  ?      Ans.  27  yr.  7  mo.  16  da.  14  h.  45  min. 

37.  How  many  cubic  yards  of  earth  were  removed  in  dig- 
ging a  cellar  28  ft.  9'  long,  22  ft.  8'  wide,  and  7  ft.  6'  deep  ? 

Ans.    ISl-^  cu.  yd. 

38.  What  will  30  bu.  54  Ib.  of  wheat  cost,  at  $1.37£  per 
bushel?  Ans.    $42.4875. 

39.  How  many  square  yards  of  carpeting  will  it  take  to 
.cover  a  floor  24  ft.  8'  long  and  18  ft.  6'  wide  ?     Ans.    50£f. 

40.  What  is  the  cost  of  54  bu.  8  Ib.  of  barley,  at  84  cents 
per  bushel  ?  Ans.    $45.50. 

41.  What  is  the  depth  of  a  lot  that  has  120  feet  front,  and 
contains  18720  square  feet? 

42.  How  many  steps  of  30  inches  each  must  a  person 
take  in  walking  21  miles? 

43.  How  long  will  it  require  one  of  the  heavenly  bodies  to 
move  through  a  quadrant,  if  it  move  at  the  rate  of  3'  12" 
per  minute  ?  Ans.    1  da.  4  h.  7  min.  30  sec. 

44.  How  many  times  will  a  wheel,  9  ft.  2  in.  in  circum- 
ference, turn  round  in  going  65  miles  ? 

45.  If  a  man   buy  10  bushels  of  chestnuts,  at  $5.00  per 
bushel,  dry  measure,  and  sell  the  same  at  22  cents  per  quart, 
liquid  measure,  how  much  is  his  gain?  Ans.    $31.92. 

46.  What  will  it  cost  to  build  a  wall  240  feet  long,  6  feet 
high,  and  3  feet  thick,  at  $3.25  per  1000  bricks,  each  brick 
being  8  inches  long,  4  inches  wide,  and  2  inches  thick  ? 

Ans.   $379.08. 


PERCENTAGE.  213 


PERCENTAGE. 

Per  cent,  is  a  term  derived  from  the  Latin  words  per 
centum,  and  signifies  by  the  hundred,  or  hundredths,  that  is,  a  cer- 
tain number  of  parts  of  each  one  hundred  parts,  of  whatever  de- 
nomination. Thus,  by  5  per  cent,  is  meant  5  cents  of  every  100 
cents,  $5  of  every  $100,  5  bushels  of  every  100  bushels,  &c. 
Therefore,  5  per  cent,  equals  5  hundredths  =.  .05  =  T^^  =  -£$. 
8  per  cent,  equals  8  hundredths  =  .08  =  T§^  =:  ^. 

232.  Percentage  is  such  a  part  of  a  number  as  is  indi- 
cated by  the  per  cent. 

233.  The  Base  of  percentage  is  the  number  on  which 
the  percentage  is  computed". 

334:.  Since  per  cent,  is  any  number  of  hundredths,  it  is 
usually  expressed  in  the  form  of  a  decimal;  but  it  may  be 
expressed  either  as  a  decimal  or  a  common  fraction,  as  in  the 

following 

TABLE. 

Decimals.        Common  Fractions.   Lowest  Terms. 


1 

per 

cent. 

m 

.01 

n= 

ToT 

= 

Too 

2 

per 

cent. 

M 

.02 

" 

ToT 

u 

-fa 

4 

per 

cent. 

II 

.04 

" 

ToT 

u 

A 

5 

per 

cent. 

u 

.05 

" 

ToT 

" 

i 

6 

per 

cent. 

U 

.06 

" 

6 

TTO" 

M 

7 

per 

cent. 

11 

.07 

" 

To~o 

" 

ToT 

8 

per 

cent. 

" 

.08 

" 

ToT 

u 

"^T 

10 

per 

cent. 

u 

.10 

u 

TTo 

" 

TO" 

16 

per 

cent. 

II 

.16 

11 

ITO 

(t 

•^ 

20 

per 

cent. 

u 

.20 

(I 

T2A 

" 

i 

25 

per 

cent. 

" 

.25 

u 

TrV 

M 

f 

50 

per 

cent. 

II 

.50 

u 

TT¥ 

M 

100 

per 

cent. 

M 

1.00 

u 

100 
too 

" 

1 

125 

per 

cent. 

" 

1.25 

(i 

m 

M 

f 

4 

per  cent.         " 

.005 

u 

1000 

U 

f 

per 

cent. 

" 

.0075 

u 

To~o  o  o 

(( 

ToT 

12* 

per 

cent. 

|| 

.125 

u 

ToTT 

u 

A 

16* 

per 

cent. 

" 

.1625 

u 

T  0  0  0  0 

u 

AMiat  is  meant  by  per  cent.  ?  From  what  is  the  term  derived  ? 
"What  is  percentage  ?  What  is  the  base  of  percentage  ?  How  is  per 
cent,  expressed? 


214  PERCENTAGE. 


EXAMPLES    FOR   PRACTICE. 


EXAMPLES    FOR    PRACTICE. 

1.    Express  decimally  3  per  cent. ;  6  per  cent. ;  9  per  cent. ; 
14  per  cent. ;  24  per  cent. ;  40  per  cent. ;  112£  per  cent. ;  150 


per  cent. 


r  cent. 

2.  Express  decimally  6J  per  cent, ;  8  j  per  cent, ;  33  J  per 
nt. ;    7£  per  cent. ;    lOf  per  cent. ;  9|  per  cent, ;  103^-  per 

cent. ;  225  per  cent. 

3.  Express  decimally  £  per  cent. ;  f  per  cent. ;  §  per  cent. ; 
f  per  cent. ;  £  per  cent. ;   1|  per  cent. ;  2|  per  cent. ;  4£  per 
cent.;  5f  per  cent.;  7|  per  cent.;   12 £  per  cent.;   25 f  per 
cent. 

4.  Express  in  the  form  of  common  fractions,  in  their  lowest 
terms,  6  per  cent. ;  8  per  cent. ;    l2  per  cent. ;   14£  per  cent. ; 
18  J  per  cent. ;    21|  per  cent, ;   31^  per  cent. ;  37£  per  cent. ; 
40  f  per  cent. ;  112  per  cent. ;  225  per  cent. 

CASE    I. 

23«>.    To  find  the  percentage  of  any  number. 

1.   A  man,  having  $125,  lost  4  per  cent,  of  it;  how  many 
dollars  did  he  lose  ?  ' 

OPERATION. 

$125  ANALYSIS.    Since  4  per  cent,  is  1^0-=  .04,  he  lost 

.04  of  $125,  or  $125  X  .04  =  $5.     Or,  4  per  cent, 
is  ^  —  Jg.,  and  -^  of  $125  =  $5.     Hence    the 


RULE.  Multiply  the  given  number  or  quantity  by  the  rate 
per  cent,  expressed  decimally,  and  point  off  as  in  decimals.  Or, 

Take  such  a  part  of  the  given  number  as  the  number  ex- 
pressing the  rate  is  part  of  100. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  6  per  cent,  of  $320  ?  Ans.    $19.20. 

3.  What  is  8  per  cent,  of  $327.25  ?  Ans.    $26.18. 

Case  I  is  what  ?     Give  explanation.    Rule. 


PERCENTAGE.  215 

4.  What  is  7%  per  cent,  of  $56.75  ?         Ans.    $4.11-^. 

5.  What  is  12J-  per  cent,  of  2450  pounds  ? 

Ans.    306.25  pounds. 

6.  What  is  6f  per  cent,  of  19072  bushels? 

Ans.   1287.36  bushels. 

7.  What  is  33£  per  cent,  of  846  gallons  ? 

Ans.   282  gallons. 

8.  What  is  9f  per  cent,  of  275  miles?     Ans.  26.95  miles- 
'9.    What  is  15  per  cent,  of  450  sheep  ? 

10.  What  is  50  per  cent,  of  1240  men  ? 

11.  What  is  105  per  cent,  of  $5760  ?  Ans.    $6048. 

12.  What  is  175  per  cent,  of  $12967  ? 

13.  What  is  25  per  cent,  of  f  ? 

25  per  cent,  equals  T2^0  =  \,  and  J  X  \  =  A»  Ans. 

14.  What  is  15  per  cent,  of  f  ?  Ans.   T^. 

15.  What  is  2£  per  cent,  of  6|  ?  Ans.   £. 

16.  What  is  33£  per  cent,  of  jk?  Ans.   •&. 

17.  What  is  84  per  cent,  of  7£?  Ans.    6^. 

18.  Find  £  per  cent,  of  $40.80  Ans.    $.306. 

19.  Find  If  per  cent,  of  $15.60  Ans.    $.26. 

20.  A  farmer,  having  760  sheep,  kept  25  per  cent,  of  them, 
and  sold  the  remainder ;  how  many  did  he  sell  ? 

21.  A  man  has  a  capital  of  $24500;    he  invests  18  per 
cent,  of  it  in  bank  stock,  30  per  cent,  of  it  in  railroad  stocks, 
and  the  remainder  in  bonds  and  mortgages ;  how  much  does 
he  invest  in  bonds  and  mortgages?  Ans.  $12740. 

22.  A  speculator  bought  1576  barrels  of  apples,  and  upon 
opening  them  he  found  12£  per  cent,  of  them  spoiled;  how 
many  barrels  did  he  lose  ? 

23.  Two  men  engaged  in  trade,  each  wfth  $2760.     One  of 
them  gained  33^  per  cent,  of  his  capital,  and  the  other  gained 
75  per  cent, ;  how  much  more  did  the  one  gain  than  the  other  ? 

Ans.    $1150. 

24.  A  man,  owning  f  of  an  iron  foundery,  sold  35  per  cent, 
of  his  share ;   what  part  of  the  whole  did  he  sell,  and  what 
part  did  he  still  own  ?  Ans.    He  still  owned  £f . 


216  PERCENTAGE. 

25.  A  owed  B  $575.40 ;  he  paid  at  one  time  40  per  cent, 
of  the  debt ;  afterward  he  paid  25  per  cent,  of  the  remainder ; 
and  at  another  time  12£  per  cent,  of  what  he  owed  after  the 
second  payment ;  how  much  of  the  debt  did  he  still  owe  ? 

Am.    $226.56f. 

CASE   II. 

236.  To  find  what  per  cent,  one  number  is  of  an- 
other. 

1.  A  man,  having  $125,  lost  $5 ;  what  per  cent,  of  his 
money  did  he  lose  ? 

OPERATION.  ANALYSIS.     We  multi- 

5  -7-  125  :=  .04  —  4  per  cent.          ply  the  base  by  the  rate 

Or,  Per  cent-   to   obtain  the 

•^  =  &  =  .04  =  4  per  cent.          Percentage  ( 235 ) ;  con- 

versely,  we  divide  the  per- 
centage by  the  base  to  obtain  the  rate  per  cent.  Or,  since  $125  is 
100  per  cent,  of  his  money,  $5  is  T|^,  equal  to  -^  of  100  per  cent, 
which  is  4  per  cent.  Hence  the 

RULE.  Divide  the  percentage  by  the  base,  and  the  quotient 
will  be  the  rate  per  cent,  expressed  decimally.  Or, 

Take  such  a  part  of  100  as  the  percentage  is  part  of  the 

base. 

EXAMPLES    FOR    PRACTICE. 

2.  What  per  cent,  of  $450  is  $90  ?  Ans.    20. 

3.  What  per  cent,  of  $1400  is  $175?  Ans.    12£. 

4.  What  per  cent,  of  $750  is  $165  ? 

f>.    What  per  cent,  of  $240  is  $13.20  ?  Ans.    5£. 

C.    What  per  cent,  of  $2  is  15  cents  ? 

7.  What  per  cent,  of  6  bushels  1  peck  is  4  bushels  2  pecks 
6  quarts  ?  Ans.    75  per  cent. 

8.  What  per  cent,  of  1 5  pounds    is   5    pounds    10   ounces 
avoirdupois  weight  ?  Ans.    37J-  per  cent. 

9.  What  per  cent,  of  250  head  of  cattle  is  40  head  ? 

Case  II  is  what  ?     Give  explanation.    Rule. 


PERCENTAGE.  217 

10.  From  a  hogshead  of  sugar  containing  760  pounds,  100 
pounds  were  sold  at  one  time,  and  90  pounds  at  another;  what 
per  cent,  of  the  whole  was  sold  ? 

11.  A  man,  having  600  acres  of  land,  sold  £  of  jt  at  one 
time,  and  ^  of  the  remainder  at  another  time ;  what  per  cent, 
remained  unsold  ?  Ans.    50  per  cent. 

CASE   III. 

237.  To  find  a  number  when  a  certain  per  cent,  of 
it  is  given. 

1.  A  man  lost  $5,  which  was  4  per  cent,  of  all  the  money 
he  had ;  how  much  had  he  at  first  ? 

OPERATION.  ANALYSIS.    We  are  here  required  to 

$5  -f-  .04  —.$125.  find  the  base,  of  which  $5  is  the  per- 

Or,  centage.     Now,  percentage  equals  base 

-5-  X  100  in  $125  multiplied  by  the  rate  per  cent;  con- 

versely, base  equals  percentage  divided 

by  rate  per  cent.  Or,  $5  is  4  per  cent,  of  all  he  had ;  \  of  $5,  or  ^, 
equals  1  per  cent,  of  all  he  had,  and  100  times  %  equals  100  per 
cent.,  or  all  he  had.  Hence  the 

RULE.  Divide  the  percentage  by  the  rate  per  cent.,  ex- 
pressed decimally,  and  the  quotient  will  be  the  base,  or  number 
required.  Or, 

Take  as  many  times  100  as  the  percentage  is  times  the  rate 
per  cent. 

EXAMPLES    FOR   PRACTICE. 

2.  16  is  8  per  cent,  of  what  number  ?  Ans.    200. 

3.  42  is  7  per  cent,  of  what  number  ? 

4.  75  is  12i  per  cent,  of  what  number?  Ans.    600. 

5.  33  is  2J  per  cent,  of  what  number  ?  Ans.    1200. 

6.  $281.25  is  374  per  cent,  of  what  sum  of  money  ? 

Ans.   $750. 

7.  A  farmer  sold  50  sheep,  which  was  20  per  cent,  of  his 
whole  flock ;  how  many  sheep  had  he  at  first  ? 

Case  III  is  what  ?     Give  explanation.     Rule. 
S 


218  PERCENTAGE. 

8.  I  loaned  a  man  a  certain  sum  of  money ;   at  one  time 
he  paid  me  $59.75,  which  was  12£  per  cent,  of  the  whole  sum 
loaned  to  him ;  how  much  did  I  loan  him  ? 

9.  A  merchant  invested  $975  in  dry  goods,  which  was  15 
per  cent,  of  his  entire  capital ;   what  was  the  amount  of  his 
capital?  Ans.    $6500. 

10.  If  a  man,  owning  40  per  cent,  of  an  iron  foundery,  sell 
25  per  cent,  of  his  share  for  $1246.50,  what  is  the  value  of 
the  whole  foundery  ?  -Ans.   $12465. 

11.  A' merchant  pays  $75  a  month  for  clerk  hire,  which  is 
25  per  cent,  of  his  entire  profits ;  how  much  are  his  profits  for 
one  year,  after  paying  his  clerk  hire  ?  Ans.   $2700. 

12.  A  produce  buyer,  having  a  quantity  of  corn,  bought 
2000  bushels  more,  and  he  found  that  this  purchase  was  40 
per  cent,  of  his  whole  stock ;   how  much  had  he  before  he 
bought  this  last  lot  ?  Ans.  3000  bushels. 


COMMISSION  AND  BROKERAGE. 

338.  An  Agent,  Factor,  or  Broker,  is  a  person  who  trans- 
acts business  for  another,  or  buys  and  sells  money,  stocks, 
notes,  &c. 

339.  Commission  is  the   percentage,   or   compensation 
allowed  an  agent,  factor,  or  commission  merchant,  for  buying 
and  selling  goods  or  produce,  collecting  money,  and  transact- 
ing other  business. 

24:0.  Brokerage  is  the  fee,  or  allowance  paid  to  a  broker 
or  dealer  in  money,  stocks,  or  bills  of  exchange,  for  making 
exchanges  of  money,  buying  and  selling  stocks,  negotiating 
bills  of  exchange,  or  transacting  other  like  business. 

NOTE.  The  rates  of  commission  and  brokerage  are  not  regulated  by 
law,  but  are  usually  reckoned  at  a  certain  per  cent,  upon  the  money 
employed  in  the  transaction. 

Define  an  agent,  factor,  or  broker.  What  is  meant  by  commission  ? 
Brokerage  ? 


COMMISSION   AND  BROKERAGE.  219 

CASE  I. 

.    To  find  the  commission  or  brokerage  011  any 
sum  of  money. 

1.  A  commission  merchant  sells  butter  and  cheese  to  the 
amount  of  $1540  ;  what  is  his  commission  at  5  per  cent.  ? 

OPERATION.  ANALYSIS. 

1540  X  .05  =  $77,  Am.  Since  the  com- 

Or,  yfo  =  2-V  ,  and  ^  X  1540  =  $77.          mission  on  &1  is 

5  cents  or  .05  of 

a  dollar,  on  $1540  it  is  $1540  X  .05  =  $77.  Or,  since  5  per  cent 
is  T|^=:^  of  the  sum  received,  the  commission  is  -fa  of  $1540 
=  $77.  Hence  the 

RULE.  Multiply  the  given  sum  by  the  rate  per  cent,  ex- 
pressed decimally,  and  the  result  will  be  the  commission  or  bro- 
kerage. Or, 

Take  such  a  part  of  the  given  sum  as  the  number  expressing 
the  per  cent,  is  part  of  100. 


EXAMPLES    FOB    PRACTICE. 

2.  A  commission  merchant  sells  goods  to  the  amount  of 
$6756  ;  what  is  his  commission  at  2  per  cent.?  Ans.  $135.12. 

3.  What  commission  must  be  paid  for  collecting  $17380, 
at  3£  per  cent.  ?  Ans.   $608.30. 

4.  An  agent  in  Chicago  purchased  4700  bushels  of  wheat, 
at  75  cents  a  bushel ;  what  was  his  commission  at  1£  per  cent, 
on  the  purchase  money  ? 

5.  A  broker  in  New  York  exchanged  $25875  on  the  Suf- 
folk Bank,  Boston,  at  £  per  cent. ;  how  much  brokerage  did 
he  receive  ?  Ans.    $64.6875. 

6.  An  auctioneer  sold  at  auction  a  house  for  $3284,  and 
the  furniture  for  $2176.50  ;  what  did  his  fees  amount  to  at 
2£  per  cent.  ? 

7.  A  broker  negotiates  a  bill  of  exchange  of  $2890  for  % 
per  cent,  commission ;  how  much  is  his  brokerage  ? 

Ans.   $23.12. 

Case  I  is  what  ?     Give  explanation.    Rule. 


220  PERCENTAGE. 

8.  An  agent  buys  for  a   manufacturing  company   26750 
pounds  of  wool,  at  32  cents  a  pound,  and  receives  a  commis- 
sion of  2  J  per  cent. ;  what  amount  does  he  receive  ? 

Ans.   $235.40. 

9.  If  I  sell  400  bales  of  cotton,  each  weighing  570  pounds, 
at  9  cents  a  pound,  and  receive  a  commission  of  2£  per  cent., 
how  much  do  I  make  by  the  transaction  ?      Ans.    $461.70. 

10.  A  commission  merchant  in  New  Orleans  sells  450  bar- 
rels of  flour  at  $7.60  a  barrel ;  38  firkins  of  butter,  each  con- 
taining 56  pounds,  at  25  cents  a  pound  ;  and  105  cheeses,  each 
weighing  48  pounds,  at   9  cents  a  pound  ;  how  much  is  his 
commission  for  selling,  at  5J  per  cent.  ?        Ans.    $243.133. 

11.  A  lawyer  collected  a  note  of  $950,  and  charged  6£  per. 
cent,  commission ;  what  was  his  fee,  and  what  the  sum  to  be 
remitted  ?  Ans.    Fee,  $61.75  ;  remitted,  $888.25. 

12.  An  insurance  agent's  fees  are  6  per  cent,  on  all  sums 
received  for  the  company,  and  4  per  cent,  additional  on  all 
sums  remaining,  at  the  end  of  the  year,  after  the  losses  are 
paid ;  he  receives,  during  the  year,  $30456.50,  and  pays  losses 
to  the  amount  of  $19814.15;  how  much  commission  does  he 
receive  during  the  year  ?  Ans.    $2253.084. 

CASE    II. 

£4£.  To  find  the  commission  or  brokerage,  when 
it  is  to  be  deducted  from  the  given  sum,  and  the  bal- 
ance invested. 

1.  A  merchant  sends  his  agent  $1260  with  which  to  buy 
merchandise,  after  deducting  his  commission  of  5  per  cent. ; 
what  is  the  sum  invested,  and  how  much  is  the  commission  ? 

OPERATION. 

$1260  ~  1.05  =  $1200,  inwKio'i. 

$1260 $1200  =  $60,  commission. 

Or,  i^  +  T&T  =  &H  $1260  -Mu-  =  $1200>  iuvested; 

And  $1260 $1200  =  $60,  commission. 


Cusc  II  is  what?     (Jive  explanation,      llulc. 


COMMISSION   AND   BROKERAGE.  221 

ANALYSIS.  Since  the  commission  is  5  per  cent.,  the  agent  must 
receive  $1.05  for  every  $1  he  expends  ;  he  can  invest  as  many 
dollars  as  $1.05  is  contained  times  in  $1260,  which  is  $1200;  and 
the  difference  between  the  given  sum  and  the  sum  invested  is  his 
commission. 

Or,  the  money  expended  is  -J$$  of  itself,  the  commission  is  T§-$  of 
this  sum,  and  the  commission  added  to  the  sum  expended  is  -j-jj-J  of 
the  whole  sum.  Since  $1260  is  f£f  =  fi$1260  -^-|i  =  $1200, 
the  sum  expended;  and  $1260  —  $1200=: $60  the  commission. 
Hence  the 

RULE.  I.  Divide  the  given  amount  by  1  increased  by  the  rate 
per  cent,  of  commission,  and  the  quotient  is  the  sum  invested. 

II.  Subtract  the  investment  from  the  given  amount,  and  the 
remainder  is  the  commission. 

EXAMPLES    FOK    PRACTICE. 

2.  A  man  sends  $3246.20  to  his  agent  in  Boston,  request- 
ing him  to  lay  it  out  in  shoes,  after  deducting  his  commission 
of  2  per  cent;  how  much  is  his  commission?    Ans.  $63.65. 

3.  What  amount  of  stock  can  be  bought  for  $9682,  and  al- 
low 3  per  cent,  brokerage  ?  Ans.    $9400. 

4.  A  flour  merchant  sent  $10246.50  to  his  agent  at  Chica- 
go, to  invest  in  flour,  after  deducting  his  commission  of  3^- 
per  cent. ;  how  many  barrels  of  flour  could  he  buy  at  $5.50 
per  barrel?  Ans.    1800  barrels. 

5.  An  agent  receives  a  remittance  of  $4908,  with  which  to 
purchase  grain,  at  a  commission  of  4^-  per  cent. ;  what  will  be 
the  amount  of  the  purchase  ? 

6.  Remitted  $603.75  to  my  agent  in  New  York,  for  the  pur- 
chase of  merchandise,  agent's  commission  being  5  per  cent. ; 
what  amount  of  broadcloth  at  $5  per  yard  should  I  receive  ? 

Ans.    115  yds. 

7.  A  commission  merchant  receives  $9376.158,  with  or- 
ders to  purchase  grain  ;  his  commission  is  3  per  cent.,  and  he 
charges  1^  per  cent,  additional  for  guaranteeing  its  delivery  at 
a  specified  time ;  how  much  will  he  pay  out,  and  what  are 
Ms  fees  ?  Ans.  Fees,  $403.758. 


222  PEP.CENTAGE. 

8.  A  real  estate  broker,  whose  stated  commission   is   1J 
per  cent.,  receives  $13842.07,  to  be  used  in  the  purchase  of 
city  lots  ;  how  much  does  he  invest,  and  what  is  his  commis- 
sion ?  Ans.    $13604  invested;  $238.07  commission. 

9.  A  broker  received  $10650,  to  be  invested  in  stocks  after 
deducting  £  per  cent,  for  brokerage ;  what  amount  of  stock 
did  he  purchase  ? 

STOCKS. 

24:3.  A  Corporation  is  a  body  authorized  by  a  general 
law,  or  by  a  special  charter,  to  transact  business  as  a  single 
individual. 

244.  A  Charter  is  the  legal  act  of  incorporation,  and  de- 
fines the  powers  and  obligations  of  the  incorporated  body. 

24:5.  A  Firm  is  the  name  under  which  an  unincorporated 
company  transacts  business. 

246.  Capital  or  Stock  is  the  property  or  labor  of  an  indi- 
vidual, corporation,  company,  or  firm;   it  receives  different 
names,  as  Bank  Stock,  Railroad  Stock,  Government  Stock,  &c. 

247.  A  Share  is  one  of  the  equal  parts  into  which  the 
stock  is  divided. 

248.  Stockholders  are  the  owners  of  the  shares. 

249.  The  Nominal  or  Par  Value  of  stock  is  its  first  cost, 
or  original  valuation. 

NOTE.  The  original  value  of  a  share  varies  in  different  companies. 
A  share  of  bank,  insurance,  railroad,  or  like  stock  is  usually  $100. 

250.  Stock  is  At  Par  when  it  sells  for  its  first  cost,  or 
original  valuation ; 

2«>1.  Above  Par,  at  a  premium,  or  in  advance,  when  it 
sells  for  more  than  its  original  cost ;  and 

252.  Below  Par,  or  at  a  discount,  when  it  sells  for  less 
than  its  original  cost. 

Define  a  corporation.  A  charter.  A  firm.  Capital  or  stock.  Shares. 
Stockholders.  Par  value.  At  par.  Above  par.  Below  par. 


STOCKS.  223 

253.  The  Market  or  Real  Value  of  stock  is  what  it  will 
bring  per  share  in  money. 

S«54r.  A  Dividend  is  a  sum  paid  to  stockholders  from  the 
profits  of  the  business  of  the  company. 

255.  An  Assessment  is  a  sum  required  of  stockholders  to 
meet  the  losses  or  expenses  of  the  business  of  the  company. 

2oG.  Premium  or  advance,  and  discount  on  stock,  divi- 
dends, and  assessments,  are  computed  at  a  certain  per  cent, 
upon  the  original  value  of  the  shares  of  the  stock. 

CASE    I. 

257.  To  find  the  value  of  stock  when  at  an  ad- 
vance, or  at  a  discount. 

1.  What  will  $3240  of  bank  stock  cost,  at  8  per  cent,  ad- 
vance ? 

OPERATION.  ANALYSIS.   Since 

§$1  -|-.08  ==  $1.08  $1  of  the  stock  at 

$3240  X  $1-08  =  $3499.20,  Am.  Par  value  wil1  cost 

$1  plus  the  premi- 
um, or  $1.08,  $3240  of  the  same  stock  will  cost  3240  X  $1.08  in 
$3499.20.  If  the  stock  w.ere  8  per  cent,  below  par,  $1  minus  the 
discount,  or  $1.00  —  $.08  =  $.92,  would  show  what  $1  of  the  stock 
would  cost.  Hence  the 

EULE.  Multiply  the  par  value  of  the  stock  by  the  number 
indicating  the  price  of  $1  of  the  same  stock,  and  the  product 
will  be  the  real  value. 

NOTE.  In  all  examples  relating  to  stocks,  $100  is  considered  the 
par  value  of  a  share  of  stock,  unless  otherwise  stated. 

EXAMPLES    FOR   PRACTICE. 

2.  If  the  stock  of  an  insurance  company  sell  at  5  per  cent. 

k below  par,  what  will  $1200  of  the  stock  cost?  Ans.  $1140. 
3.   What  is  the  market  value  of  35  shares  of  New  York 
Central  Railroad  stock,  at  15  per  cent,  below  par  ? 
Market  value.     A  dividend.    An  assessment.    Case  I  is  what?     Give 
explanation.     Rule. 


224  PERCENTAGE. 

4.  What  must  be  paid  for  48  shares  of  Panama  Railroad 
stock,  at  a  premium  of  5^  per  cent.,  if  the  par  value  be  $150 
per  share?  A/is.    $7596. 

5.  What  costs  $5364  stock  in  the  Minnesota  copper  mines, 
at  9  per  cent,  above  par  ? 

6.  A  man  purchased  $6275  stock  in  the  Pennsylvania  Coal 
Company,  and  sold  the  same  at  a  discount  of  12  per  cent.; 
what  was  his  loss  ?  Ans.    $753. 

7.  What  must  be  paid  for   125   shares  of  United   States 
stock,  at  4J  per  cent,  premium,  the  par  value  being  $1000  per 
share?  Ans.    $130937.50. 

8.  Bought  42  shares  of  Illinois  Central  Railroad  stock,  at 
14  per  cent,  discount,  and  sold  the  same  at  an  advance  of  12£ 
per  cent. ;  how  much  did  I  gain  ?  Ans.    $1113. 

9.  What  is  the  market  value  of  175  shares  of  stock  in  the 
Suffolk  Bank,  at  J  per  cent,  advance  ?        Ans.    $17631.25. 

10.  Bought  75  shares  of  stock  in  the  Bank  of  New  Orleans, 
of  $50  each,  at  3  per  cent,  discount,  and  sold  it  at  2£  per  cent, 
advance;  what  was  my  gain  ?  Ans.    $196.875. 

11.  B  exchanged  28  shares  of  bank  stock,  of  $50  each, 
worth  7  per  cent,  premium,  for  25  shares  of  railroad  stock,  of 
$100  each,  at  12£  per  cent,  discount,  and  paid  the  difference 
in  cash ;  how  much  cash  did  he  pay  ?  Ans.    $689.50. 

CASE    II. 

358.   To  find  how  much  stock  may  be  purchased  for 
a  given  sum. 

1.    How  many  shares  of  bank  stock,  at  3  per  cent,  advance, 
may  be  bought  for  $5150  ? 

OPERATION.  ANALYSIS.    Since  the  stock 

$5150  -^-  1.03  =  $5000  =          is  at  3  per  cent,  advance,  $1 

50  shares    Ans.  °^  stock  at  par  will  cost  $1.03 ; 

and  if  we  divide   $5150,  the 

whole  sum  to  be  expended,  by  $1.03,  the  cost  of  $1  of  stock,  the 
quotient  must  be  the  amount  of  stock  purchased.     Hence  the 

Case  n  is  what  ?     Give  explanation. 


PROFIT  AND  LOSS.  225 

RULE.     Divide  the  given  sum  by  the  cost  of  $1  of  stock, 
and  the  quotient  will  be  the  nominal  amount  of  stock  purchased. 

2.  How  many  shares  of  railroad  stock,  at  5  per  cent,  ad- 
vance, can  be  purchased  for  $6300  ?  Ans.    60  shares. 

3.  I  invested  $6187.50,  in  Ocean  Telegraph  stock,  at  10 
per  cent,  discount  ;  how  much  stock  did  I  purchase  ? 

Ans.   $6875. 

4.  I  sent  my  agent  $53500  to  be  invested  in  Illinois  Cen- 
tral Railroad  stock,  which  sold  at  7  per  cent,  advance  ;  what 
amount  did  he  purchase  ?  Ans.    $50000. 

5.  Sold  50  shares  of  stock  in  a  Pittsburg  ferry  company, 
at  8  per  cent,  discount,  and  received  $1150;  what  is  the  par 
value  of  1  share  ?  Ans.   $25. 

PROFIT  AND  LOSS. 

259.  Profit  and  Loss  are  commercial  terms,  used  to  ex- 
press the  gain  or  loss  in  business  transactions,  which  is  usually 

'  reckoned  at  a  certain  per  cent,  on  the  prime  or  first  cost  of 
articles. 

CASE   I. 

260.  To  find  the  amount  of  profit  or  loss,  when  the 
cost  and  the  gain  or  loss  per  cent,  are  given. 

1.   A  man  bought  a  horse  for  $135,  and  afterward  sold  him 
for  20  per  cent,  more  than  he  gave  ;  how  much  did  he  gain  ? 
OPERATION.  ANALYSIS.      Since    $1 

$135  X  .20=  $27,  Ans.  Sains  2<>  cents,  or  20  per 

i     <ttiQ*vi  _  <£97         cent,  $135  will  gain  $135 
=  *;  $    o  X  *  -  »27. 


>  since  20 

per  cent,  equals  -££$  =  ^,  the  whole  gain  will  be  ^  of  the  cost. 
Hence  the  following 

RULE.  Multiply  the  cost  by  the  rate  per  cent,  expressed 
decimally.  Or, 

Take  such  part  of  the  cost  as  the  rate  per  cent,  is  part  of  100. 


Rule.     What  is  meant  by  profit  and  loss  ?     Case  I  is  what  ?     Give 
explanation.    Rule. 


226  PERCENTAGE. 

EXAMPLES    FOB   PRACTICE. 

2.  A  grocer  bought  a  hogshead  of  sugar  for  $84.80,  and  sold 
it  at  12J-  per  cent,  profit ;  what  was  his  gain  ? 

3.  A  miller  bought  500  bushels  of  wheat  at  $1.15  a  bushel, 
and  he  sold  the  flour  at  16§  per  cent,  advance  on  the  cost  of 
the  wheat;  what  was  his  gain?  Ans.  $95.834-. 

4.  Bought  76  cords  of  wood  at  $3.62£  a  cord,  and  sold  it 
so  as  to  gain  26  per  cent. ;  what  did  I  make  ? 

5.  A  hatter  bought  40  hats  at  $1.75  apiece,  and  sold  them 
at  a  loss  of  14f-  per  cent. ;  what  was  his  whole  loss  ? 

6.  A  grocer  bought  3  barrels  of  sugar,  each  containing  230 
pounds,  at  8£  cents  a  pound,  and  sold  it  at  18T2T  per  cent,  profit ; 
what  was  his  whole  gain,  and  what  the  selling  price  per  pound  ? 

Ans.  Whole  gain,  $10.35  ;  price  per  pound,  9£  cents. 

7.  A  sloop,  freighted  with  3840  bushels  of  corn,  encoun- 
tered a  storm,  when  it  was  found  necessary  to  throw  37^  per 
cent,  of  her  cargo  overboard ;  what  was  the  loss,  at  62£  cents 
a  bushel  ?  Ans.    $900  loss. 

8.  A  gentleman  bought  a  store  and  contents  for  $4720 ;  he 
sold  the  same  for  12J-  per  cent,  less  than  he  gave,  arid  then 
lost  15  per  cent,  of  the  remainder  in  bad  debts ;  what  was  his 
entire  loss  ?  Ans.  $1209.50. 

9.  A  man  commenced  business  with    $3000  capital ;    the 
first  year  he  gained  22£  per  cent.,  which  he  added  to  his  capi- 
tal ;  the  second  year  he  gained  30  per  cent,  on  the  whole  sum, 
which  gain  he  also  put  into  his  business ;  the  third  year  he 
lost  1 6§  per  cent,  of  his  entire  capital ;  how  much  did  he  make 
in  the  3  years  ?  Ans.    $981.25. 

CASE    II. 

261.    To  find  the  gain  or  loss  per  cent.,  when  the 
cost  and  selling  price  are  given. 

1.    Bought  wool  at  32  cents  a  pound,  and  sold  it  for  40  cents 
a  pound  ;  what  per  cent,  was  gained  ? 

Case  II  is  what  ?     Give  explanation.    Rule. 


PROFIT  AND   LOSS.  227 

OPERATION. 

40  _  32  —  8  ;  8  -j-  32  =  &  —  .25,  ^W5. 
6r,  40  —  32  =  8 ;  8  ^-  32  =  -fa  =  £;  £  X  100  =  25  per  cent. 

ANALYSIS.  Since  the  gain  on  32  cents  is  40  —  32  =:  8  cents,  the 
whole  gain  is  -£%  =.  -J-  of  the  purchase  money ;  and  ^  reduced  to  a 
decimal  is  25  hundredths,  equal  to  25  per  cent.  Or,  if  the  gain  were 
equal  to  the  purchase  money,  it  would  be  100  per  cent. ;  but  since 
the  gain  is  -^  =  £  of  the  purchase  money,  it  will  be  \  of  100  per 
cent.,  equal  to  25  per  cent.  Hence  the  following 

RULE.  Make  the  difference  between  the  purchase  and  selling 
prices  the  numerator,  and  the  purchase  price  the  denominator  ; 
reduce  to  a  decimal,  and  the  result  will  be  the  per  cent.  Or, 

Take  such  a  part  of  100  as  the  gain  or  loss  is  part  of  the 
purchase  price. 

EXAMPLES    FOR    PRACTICE. 

2.  A  man  bought  a  pair  of  horses  for  $275,  and  sold  them 
for  $330  ;  what  per  cent,  did  he  gain  ?     Ans.    20  per  cent. 

3.  If  a  merchant  buy  cloth  at  $.60  a  yard,  and  sell  it  for 
$.75  a  yard,  what  does  he  gain  per  cent.  ? 

4.  A  speculator  bought  108  barrels  of  flour  at  $4.62£  a 
barrel,  and  sold  it  so  as  to  gain  $114.88^-;  what  per  cent, 
profit  did  he  make  ?  Ans.    23  per  cent. 

5.  Bought  sugar  at  8  cents  a  pound,  and  sold  it  for  9£  cents 
a  pound ;  what  per  cent,  was  gained  ? 

6.  A  drover  bought  150  head  of  cattle  for  $42  per  head, 
and  sold  them  for  $5400  ;  what  was  his  loss  per  cent.  ? 

Ans.    14f  per  cent. 

7.  If  I  sell  for  $15  what  cost  me  $25,  what  do  I  lose  per 
cent.  ?  Ans.   40  per  cent. 

8.  Bought  paper  at  $2  per  ream,  and  sold  it  at  25  cents 
a  quire;  what  was  the  gain  per  cent.  ?    Ans.  150  per  cent. 

9.  If  I  sell  £  of  an  article  for  f  of  its  cost,  what  is  gained 
per  cent.  ?  Ans.    50  per  cent. 

10.  If  ^  of  an  article  be  sold  for  what  £  of  it  cost,  what  is 
the  loss  per  cent.  ?  Ans.  37£  per  cent. 


228  PERCENTAGE. 

11.  If  I  sell  3  pecks  of  clover-seed  for  what  one  bushel 
cost  me,  what  per  cent,  do  I  gain  ?  Ans.    33£  per  cent. 

12.  A,  having  a  debt  against  B,  agreed  to  take  $.871  on 
the  dollar  ;  what  per  cent,  did  A  lose  ? 

13.  A  grocer  bought  7  cwt.  20  Ib.  of  sugar,  at  7  cents  a 
pound,  and  sold  3  cwt.  42  Ib.  at  8  cents,  and  the  remainder  at 
8^-  cents  ;  what  was  his  gain  per  cent.  ?  Ans.  18^  per  cent. 

14.  Bought  2  hogsheads   of  wine,  at  $1.25  a  gallon,  and 
sold  the  same  at  $1.60  ;  what  was  the  whole  gain,  and  what 
the  gain  per  cent.  ?  Ans.    Gain  28  per  cent. 

15.  A  grain  dealer  bought  corn  at  $.55  a  bushel  and  sold 
it  at  $.66,  and  wheat  for  $1.10,  and  sold  it  for  $1.37^-;  upon 
which  did  he  make  the  greater  per  cent.  ? 

Ans.    o  per  cent.,  upon  the  wheat. 

CASE    III. 

263.    To  find  the  selling  price,  when  the  cost  and 
the  gain  or  loss  per  cent,  are  given. 

1.    Bought  a  horse  for  $136  ;  for  how  much  must  he  be  sold 
to  gain  25  per  cent.  ? 

OPERATION.  ANALYSIS.    Since  $1  of  cost 

$1  +  .25  =z  $1.25.  sells  for  $1.25,  $136  of  cost  will 

$1.25  X  136  =  $170,  Ans.  sel1  for  136  timcs  $!-25>  which 

Or    i  «£  -4-   25    —  J.25  _  5  equals  $170,  the  selling  price. 

~  °r'  slnce  the  cost  is  «*>  and 


will  be  l-f  £  =  }  of  the  cost,  or 

}  of  $136  =  $170.  If  the  horse  had  been  sold  at  a  loss  of  25  per 
cent.,  then  $1  of  cost  would  have  sold  for  $1  minus  .25,  or  $.75, 
&c.  Hence, 

RULE.  Multiply  $1  increased  by  the  gain  or  diminished  Inj 
the  loss  per  cent,  by  the  number  denoting  the  cost.  Or, 

Take  such  a  part  of  the  cost  as  is  equal  to  -f{}$  increased  or 
diminished  by  the  gain  or  loss  per  cent. 

Case  ILI  is  what  ?    Give  explanation.    Rule. 


PKOFIT  AND  LOSS.  229 

EXAMPLES    FOR   PRACTICE. 

2.  If  12£  hundred  weight  of  sugar  cost  $140,  how  must 
it  be  sold  per  pound  to  gain  25  per  cent.  ?  Ans.  14  cents. 

o.  Bought  a  hogshead  of  molasses  for  30  cents  a  gallon, 
and  paid  16f  per  cent,  on  the  prime  cost,  for  freight  and  cart- 
age ;  how  much  must  it  sell  for,  per  gallon,  to  gain  33^-  per 
cent,  on  the  whole  cost  ?  Ans.  $.46§. 

4.  For  what  price  must  I  sell  coffee  that  cost  10^  cents  a 
pound,  to  gain  17£  per  cent.? 

5.  If  I  am  compelled  to  sell  damaged  goods  at  a  loss  of  15 
per  cent.,  how  should  I  mark    goods  that  cost  me   $.62^-  ? 

»$1.20?  $3.87^- ?  Ans.    $.53£;  $1.02;  $3.29f. 

6.  A  man,  wishing  to  raise  some  money,  offers  his  house 
and  lot,  which  cost  him  $3240,  for  18  per  cent,  less  than  cost ; 
what  is  the  price  ?  % 

7.  C  bought  a  farm  of  120  acres,  at  $28  an  acre,  paid 
$480  for  fencing,  and  then  sold  it  for  12^  per  cent,  advance 
on  the  whole  cost ;  what  was  his  whole  gain,  and  what  did  he 
receive  an  acre  ?  Ans.    $480  gain  ;  $36  an  acre. 

8.  Bought   a   cask   of  brandy,  containing  52   gallons,  at 
$2.60  per  gallon  ;  if  7  gallons  leak  out,  how  must  the  remain- 
der be  sold  per  gallon,  to  gain  37J-  per  cent,  on  the  cost  of  the 
whole?  Ans.    $4.13£. 

9.  A  merchant  bought  15  pieces  of  broadcloth,  each  piece 
containing  23^  yards,  for  $840,  and  sold  it  so  as  to  gain  18J 
per  cent. ;  how  much  did  he  receive  a  yard  ? 

CASE    IV. 

263.  To  find  the  cost,  when  the  selling  price  and 
the  gain  or  loss  per  cent,  are  given. 

tl.   A  merchant  sold  cloth  for  $4.80  a  yard,  and  by  so  doing 
lade  33^  per  cent. ;  how  much  did  it  cost  ? 
OPERATION. 
$1  _j_  .331.  =  $l.33i ;  $4.80  -r-'l.33£  =  $3.60,  Ans. 
Or,  $4.80  —  |  of  the  cost ;  $4.80  -=-  £  =  $3.60. 
T  Case  IV  is  what? 


230  PERCENTAGE. 

ANALYSIS.  Since  the  gain  is  33^-  per  cent,  of  the  cost,  $1  of  the 
cost,  increased  by  33|-  per  cent.,  will  be  what  $1  of  cost  sold  for : 
therefore  there  will  be  as  many  dollars  of  cost,  as  1.33^  is  con- 
tained times  in  $4.80,  or  $3.60.  Or,  since  he  gained  33£  per  cent. 
=  £  of  the  cost,  $4.80  is  £  of  the  cost ;  $4.80  ~-  £  =  $3.60. 

NOTE.  If  the  rate  per  cent,  be  loss,  we  subtract  it  from  $1,  instead 
of  adding  it.  Hence  the  following 

RULE.  Divide  the  selling  price  by  $1  increased  by  the  gain 
or  diminished  by  the  loss  per  cent.,  expressed  decimally,  or  in 
the  form  of  a  common  fraction,  and  the  quotient  will  be  the 
cost. 

EXAMPLES   FOR   PRACTICE. 

2.  By  selling  sugar  at  8  cents  a  pound,  a  merchant  lost  20 
per  cent. ;  what  did  the  sugar  cost  him  ?        Ans.    10  cents. 

3.  Sold  flour  for  $6.12£  per  barrel,  and  lost  12£  per  cent. ; 
what^was  the  cost  ?  Ans.    $7.00. 

4.  A  grocer,  by  selling  tea  at  $.96  a  pound,  gains  28  per 
cent.  ;  how  much  did  it  cost  him  ?  Ans.    $.75. 

'5.  Sold  a  quantity  of  flour  for  $1881,  which  was  18J  per 
cent,  more  than  it  cost ;  how  much  did  it  cost  ? 

6.  Sold  25  barrels  of  apples  for  $69.75,  and  made  24  per 
cent. ;  how  much  did  they  cost  per  barrel  ? 

7.  Sold  9£  cwt.  of  sugar  at  $8J  per  cwt.,  and  thereby  lost 
12  per  cent. ;  how  much  was  the  whole  cost? 

8.  Having  used  a  carriage  six  months,  I  sold  it  for  $96, 
which  was  20  per  cent,  below  cost ;  what  would  I  have  received 
had  I  sold  it  for  15  per  cent. above  cost?  Ans.    $138. 

9.  B  sells  a  pair  of  horses  to  C,  and  gains  12£  per  cent. ; 
C  sells  them  to  D  for  $570,  and  by  so  doing  gains  18f  per 
cent. ;  how  much  did  the  horses  cost  B  ?      Ans.    $426.66f . 

10.  A  grocer  sold  4  barrels  of  sugar  for  $24  each  ;  on  2 
barrels  he  gained  20  per  cent.,  and  on  the  other  2  he  lost  20 
per  cent. ;  did  he  gain  or  lose  on  the  whole  ?     Ans.   Lost  $4. 

11.  A  person  sold  out  his  interest  in  business  for  $4900, 
which  was  40  per  cent,  iriore  than  3  times  as  much  as  he  began 
with ;  how  much  did  he  begin  with  ?  Ans.    $1166.66|. 

Give  explanation.    Rule. 


INSURANCE.  231 


INSURANCE. 

264.  Insurance  on  property  is  security  guaranteed  by 
one  party  to  another,  for  a  stipulated  sum,  against  the  loss  of 
that  property  by  fire,  navigation,  or  any  other  casualty. 

265.  The  Insurer  or  Underwriter  is  the  party  taking  the 
risk. 

266.  The  Insured  is  the  party  protected. 

267.  The   Policy  is  the  written   contract   between   the 
parties. 

268.  The  Premium  is  the  sum  paid  by  the  insured  to  the 
insurer,  and  is  estimated  at  a  certain  rate  per  cent,  of  the 
amount  insured,  which  rate  varies  according  to  the  degree  of 
hazard,  or  class  of  risk. 

XOTE.  As  a  security  against  fraud,  most  insurance  companies  take 
risks  at  not  more  than  two  thirds  the  full  value  of  the  property 
insured. 

269*  To  find  the  premium  when  the  rate  of  insur- 
ance and  the  amount  insured  are  given. 

1.  What  must  I  pay  annually  for  insuring  my  house  to  the 
amount  of  $3250,  at  1£  per  cent,  premium  ? 

OPERATION.  ANALYSIS.      We 

$3250  X  -01£  or  .0125  =  $40.625.          multiply  the  amount 

Or,  1£  per  ct.  =  jfo  =  &  ;  insured,    $3250,   by 

$3250  X  A  =  &40.62A.  the   ra*e!  **  P?r 

cent.,  and  the  result, 

$40.625,  is  the  premium.     Or,  the  rate,  \\  per  cent,  is  -^=.-fa  of 
the  amount  insured,  and  -fa  of  $3250  is  $40.62^.     Hence  the 

RULE.  Multiply  the  amount  insured  by  the  rate  per  cent., 
and  the  product  will  be  the  premium.  Or, 

Take  such  a  part  of  the  amount  insured  as  the  rate  is  part 
of  100. 

Define  insurance.  Insurer,  or  underwriter.  Policy.  Premium. 
To  what  amount  can  property  usually  be  insured  ?  Give  analysis  of 
example  1.  Rule. 


232  PERCENTAGE. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  premium  on  a  policy  for  $750,  at  4  per 
cent.?  Ans.    $30. 

3.  What  premium  must  be  paid  for  $4572.80  insurance,  at 
2£  percent.?  Ans.    $114.32. 

4.  A  house  and  furniture,  valued  at  $5700,  are  insured  at 
If  per  cent. ;  what  is  the  premium  ?  Ans.    $99.75. 

5.  A  vessel  and  cargo,  valued  at  $28400,  are  insured  at  3£ 
per  cent.;  what  is  the  premium?  Ans.    $994. 

6.  A  woolen  factory  and  contents,  valued  at  $55800,  are 
insured  at  2%  per  cent. ;   if  destroyed  by  fire,  what  would  be 
the  actual  loss  of  the  company  ?  Ans.    $54237.60. 

7.  What  must   be  paid  to  insure  a  steamboat  and  cargo 
from  Pittsburg  to  New  Orleans,  valued  at  $47500,  at  f  of  1 
percent.?  Ans.    $356.25. 

8.  A  gentleman  has  a  house,  insured  for  $8000,  and  the  fur- 
niture for  $4000,  at  2f  per  cent. ;   what  premium  must  he 
pay?  Ans.   $285. 

9.  A  cargo  of  4000  bushels  of  wheat,  worth  $1.20  a  bushel, 
is  insured  at  J  of  1^-  per  cent,  on  f  of  its  value ;  if  the  cargo  be 
lost,  how  much  will  the  owner  of  the  wheat  lose  ?    Ans.   $1636. 

10.  What  will  it  cost  to  insure  a  factory  valued  at  $21000, 
at  |  per  cent. ;  and  the  machinery  valued  at  $15400,  at  |  per 
cent.?  Ans.   $264.25. 

TAXES. 

270.  A  Tax  is  a  sum  of  money  assessed  on  the  person 
or  property  of  an  individual,  for  public  purposes. 

271.  When  a  tax  is  assessed  on  property,  it  is  apportioned 
at  a  certain  per  cent,  on  the  estimated  value. 

When  assessed  on  the  person,  it  is  apportioned  equally 
among  the  male  citizens  liable  to  assessment,  and  is  called  a 
poll  tax.  Each  person  so  assessed  is  called  a  poll. 

What  is  a  tax  r  How  is  a  tax  on  property  apportioned  ?  On  the 
person,  how  ? 


TAXES.  233 

S72J.  Property  is  of  two  kinds  —  real  estate,  and  personal 
property. 

S73.  Real  Estate  consists  of  immovable  property,  such 
as  lands,  houses,  &c. 

££74.  Personal  Property  consists  of  movable  property, 
such  a,s  money,  notes,  furniture,  cattle,  tools,  &c. 

£175.  An  Inventory  is  a  written  list  of  articles  of  proper- 
ty, with  their  value. 

S76 .  Before  taxes  are  assessed,  a  complete  inventory  of  all 
the  taxable  property  upon  which  the  tax  is  to  be  levied  must 
be  made.  If  the  assessment  include  a  poll  tax,  then  a  complete 
list  of  taxable  polls  must  also  be  made  out. 

I.  A  tax  of  $3165  is  to  be  assessed  on  a  certain  town; 
the   valuation  of  the  taxable  property,  as  shown  by  the  as- 
sessment roll,  is  $600,000,  and  there  are  220  polls  to  be  as- 
sessed 75  cents  each ;  what  will  be  the  tax  on  a  dollar,  and 
how  much  will  be  A's  tax,  whose  property  is  valued  at  $3750, 
and  who  pays  for  3  polls? 

OPERATION. 

$.75  X  220  =  $165,  amount  assessed  on  the  polls. 

$3165  —$165  =  $3000,  amount  to  be  assessed  on  the  property. 

$3000  -J-  $600,000  =  .005,  tax  on  $1. 

$3750  X  -005  =  $18.75,  A's  tax  on  property. 

$.75  X  3  =  $2.25,  A's  tax  on  3  polls. 

$18.75  4-  $2.25  =  $21,  amount  of  A's  tax. 

Hence  the  following 

RULE.  I.  Find  the  amount  of  poll  tax,  if  any,*  and  subtract 
this  sum  from  the  whole  amount  of  tax  to  be  assessed. 

II.  Divide  the  sum  to  be  raised  on  property,  by  the  whole 
amount  of  taxable  property,  and  the  quotient  will  be  the  per 
cent.,  or  the  tax  on  one  dollar. 

III.  Multiply  each  man's  taxable  property  by  the  per  cent., 
or  the  tax  on  $1,  and  to  the  product  add  his  poll  tax,  if  any  ; 
the  result  will  be  the  whole  amount  of  his  tax. 

What  is  real  estate  ?  Personal  property  ?  An  inventory  ?  Explain 
the  process  of  levying  a  state  or  other  tax.  Rule. 


234 


PERCENTAGE. 


NOTE.  Having  found  the  tax  on  $1,  or  the  per  cent.,  which  in  the 
preceding  example  we  find  to  be  5  mills,  or  ^  per  cent.,  the  operation 
of  assessing  taxes  may  be  greatly  facilitated  by  finding  the  tax  011  $2, 
$3,  &c.,  to  $10,  and  then  on  $20,  $30,  &c.,  to  $100,  and  arranging 
the  numbers  as  in  the  following 


TABLE. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1  gives 

$.005 

$10 

$.05 

$100 

$  .50 

$1000 

$5.00 

2    » 

.01 

20 

.10 

200 

1.00 

2000 

10. 

3    « 

.015 

30 

.15 

300 

1.50 

3000 

15. 

4    " 

.02 

40 

.20 

400 

2.00 

4000 

20. 

5    » 

.025 

50 

.25 

500 

2.50 

5000 

25. 

6    « 

.03 

60 

.30 

600 

3.00 

6000 

30. 

7    « 

.035 

70 

.35 

700 

3.50 

7000 

35. 

8    « 

.04 

80 

.40 

800 

4.00 

8000 

40. 

9    " 

.045 

90 

.45 

900 

4.50 

9000 

45. 

EXAMPLES    FOR   PRACTICE. 

2.  According  to  the  conditions  of  the  last  example,  how 
much  would  be  a  person's  tax  whose  property  was  assessed 
at  $3845,  and  who  paid  for  2  polls  ? 

Finding  the  amount  from  the  table, 

The  tax  on  $3000 is  .  . 

"       "     "         800  .  .  ,  .  " 


5  .  .     .  " 
2  polls  " 


$15.00 
4.00 
.20 
.025 
1.50 


Total  tax  is $20.725 

3.  How  much  would  be  Ws  tax,  who  was  assessed  for  1 
poll,  and  on  property  valued  at  $5390  ?  Am.    $27.70.  • 

4.  A  tax  of  $9190.50  is  to  be  assessed  on  a  certain  village  ; 
the  property  is  valued  at  $1400000,  and  there  are  2981  polls, 
to  be  taxed  50  cents  each  ;  what  is  the  assessment  on  a  dollar  ? 
what  is  C's  tax,  his  property  being  assessed  at  $12450,  and  he 
paying  for  2  polls?     Am.  $.005J  on  $1 ;   $('»'.).  17 i,  C's  lax. 

5.  What  is  the  tax  of  a  non-resident,  having  property  in 
the  same  village  valued  at  $5375  ?  Ans.    $29.5625. 


Explain  the  table  and  its  use. 


CUSTOM    HOUSE   BUSINESS.  235 

6.  A  mining   corporation,    consisting   of  30   persons,  are 
taxed  $4342.75 ;  their  property  is  assessed  for  $188000,  and 
each  poll  is  assessed  62^-  cents ;  what  per  cent,  is  their  tax, 
and  ho\v  much  must  he  pay  whose  share  is  assessed  for  $2500, 
and  who  pays  for  1  poll  ?  Ans.  2T3,y  per  cent.;  $58.125. 

7.  In  a  certain  county,  containing   25482   taxable  inhab- 
itants, a  tax  of  $103294.60  is  assessed  for  town,  county,  and 
state  purposes ;  a  part  of  this  sum  is  raised  by  a  tax  of  30 
cents  on  each  poll ;  the  entire  valuation  of  property  on  the  as- 
sessment roll  is  $38260000  ;  what  per  cent,  is  the  tax,  and  how 
much  will  a  person's  tax  be  who  pays  for  3  polls,  and  whose 
property  is  valued  at  $9470  ?  Ans.  to  last,  $24.575. 

8.  The  number  of  polls  in  a  certain  school  district  is  225, 
and  the  taxable  property  $1246093.75  ;  it  is  proposed  to  build 
a  union  school  house  at  an  expense  of  $10000;  if  the  poll  tax 
be  $1.25  a  poll,  and  the  cost  of  collecting  be  2^-  per  cent.,  what 
will  be  the  tax  on  a  dollar,  and  how  much  will  be  E's  tax,  who 
pays  for  1  poll,  and  has  property  to  the  amount  of  $11500  ? 

Ans.    $.008,  tax  on  $1 ;  $93.25,  E's  tax. 

9.  In  a  certain  district  the  school  was  supported  by  a  rate- 
bill  ;  the  teacher's  wages  amounted  to  $300,  the  fuel  and  other 
expenses  to  $50.50 ;  the  public  money  received  was  $92,  and 
the  whole  number  of  days'  attendance  was  2585 ;   A  sent  2 
pupils  118  days  each ;  how  much  was  his  rate-bill  ?  Ans.  $2.36. 


CUSTOM  HOUSE  BUSINESS. 

S77.  Duties,  or  Customs,  are  taxes  levied  by  government 
on  imported  goods,  for  the  support  of  government  and  the 
protection  of  home  industry. 

278.  A  Custom  House  is  an  office  established  by  govern- 
ment for  the  transaction  of  business  relating  to  duties. 

It  is  lawful  to  introduce  merchandise  into  a  country  only 

Define  duties.    A  custom  house. 


236  PERCENTAGE. 

at  points  where  custom  houses  are  established.  A  seaport 
town,  having-  a  custom  house,  is  called  a  port  of  entry.  To 
carry  on  foreign  commerce  secretly,  without  paying  the  duties 
imposed  by  law,  is  smuggling. 

27O.  Tonnage  is  a  tax  levied  upon  the  vessel,  independent 
of  its  cargo,  for  the  privilege  of  coming  into  a  port  of  entry. 
Its  amount  is  regulated  by  the  size  of  the  vessel. 

28O.  Revenue  is  the  income  to  government  from  duties 
and  tonnage. 

Duties  are  of  two  kinds  —  ad  valorem  and  specific. 
.  Ad  Valorem  Duty  is  a  percentage  computed  on  the 
market  value  of  merchandise  in  the  country  from  which  it  is 
imported. 

383.  Specific  Duty  is  a  sum  computed  on  the  weight  or 
measure  of  goods,  without  regard  to  their  cost. 

CASE    I. 

384.  To  compute  ad  valorem  duties. 

1.  What  is  the  duty,  at  8  per  cent.,  on  a  quantity  of  silk 
which  cost,  in  Florence,  $6850  ? 

OPERATION.  ANALYSIS.     Since  the  duty  is  ad  valorem, 

$6850  we  multiply  the  cost,  $6850,  by  .08,  the 

.08  given  rate  per  cent.,  and  we  have  $548,  the 

Ans.    Hence  the 
$548.00,  Ans. 

RULE.  Multiply  the  cost  of  the  goods  ly  the  given  rate  per 
cent.,  and  the  product  will  be  the  required  duty. 

EXAMPLES    FOR   PRACTICE. 

2.  What  is  the  duty,  at  6  per  cent.,  on  a  shipment  of  spring 
hats,  bought  in  Paris  for  $8565  ?  Ans.    $513.90. 

3.  John  Jones  imported,  from  Havana,  50  hhds.  of  W.  I. 
molasses,  which  were  invoiced  at  36  cents  per  gallon ;  what 
was  the  duty  at  15  per  cent.  ?  Ans.    $170.10. 

Define  a  port  of  entry.  Smuggling.  Tonnage.  Revenue.  Ad 
valorem  duty.  Specific  duty.  What  is  Case  I  ?  Give  explanation. 
Rule. 


CUSTOM   HOUSE  BUSINESS.  287 

4.  What  is  the  duty,  at  12  per  cent.,  on  175  bags  of  Java 
coffee,  each  containing  115  Ibs.,  valued  at  15  cents  per  pound  ? 

Ans.   $362.25. 

CASE    II. 

285.  To  compute  specific  duties. 

It  is  the  design  of  government  to  tax  only  so  much  of  the 
merchandise  as  will  be  available  to  the  importer  in  market ; 
consequently,  the  law  directs  that  certain  allowances  be  made 
before  specific  duties  are  computed. 

286.  Gross  Weight  is  the  weight  of  the  goods  together 
with  the  package  or  covering  which  contains  them. 

287.  Net  Weight  is  what  remains  after  all  allowances 
have  been  made. 

288.  Draft,  or  Tret,  is  an  allowance  for  waste,  and  is  to  be 
deducted  always  from  the  gross  weight  or  measure. 

DKAFT   BY   WEIGHT. 

On        112  Ib 1  lb. 

Above  112  lb.    and  not  exceeding    224  lb.,  2  lb. 

"       224  lb.      "      "  "  336  lb.,  3  lb. 

«       336  lb.      "      "  "          1120  lb.,  4  lb. 

"     1120  lb.      "      "  "          2016  lb.,  7  lb. 

«     2016  lb 9  lb. 

By  law  and  usage  no  greater  allowance  than  9  lb.  can  be 
made  for  draft  on  one  box  or  package ;  and  sometimes  several 
pieces  are  weighed  together  for  one  draft. 

289.  Leakage   is   an   allowance   of  2   per  cent,  on   all 
liquors  in  casks  paying  duty  by  the  gallon. 

290.  Breakage  is  an  allowance  of  10  per  cent,  on  ale, 
beer,  and  porter,  in  bottles ;  and  of  5  per  cent,  on  all  other 
liquors  in  bottles. 

NOTE.  Leakage  and  breakage  are  the  principal  kinds  of  draft 
allowed  for  liquors. 

291.  Tare  is  an  allowance  for  the  weight  of  the  pack- 

What  is  Case  II  ?  Define  gross  weight.  Net  weight.  Draft.  Leak- 
age.  Breakage.  Tare. 


Zoo  PERCENTAGE. 

age  or  covering  that  contains  the  goods.  It  is  sometimes 
reckoned  by  the  piece,  but  is  generally  a  certain  per  cent,  of 
that  part  of  the  gross  weight  which  remains  after  the  deduc- 
tion of  draft. 

To  compute  ad  valorem  duties,  we  have  the  following 

RULE.  I.  When  the  merchandise  is  subject  to  leakage  or 
breakage; —  Deduct  the  legal  alloivance,  and  compute  the  duty 
on  the  remainder. 

II.  When  the  merchandise  is  subject  to  both  draft  and 
tare; — First  deduct  the  draft,  and  on  the  remainder  compute 
the  tare,  and  deduct  this  also  ;  then  compute  the  duty  on  the 
net  weight  or  measure. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  duty,  at  2  cents  per  pound,  on  6  barrels  of 
sugar,  weighing  250  pounds  apiece,  allowing  draft  for  each 
barrel  separately,  and  tare  at  15  per  cent.  ?     Ans.  $25.194. 

2.  What  is  the  duty  on  50  casks  of  nails,  weighing  each 
450  Ibs.  gross,  at  1^  cents  per  pound,  draft  being  allowed  for 
each  cask,  and  tare  at  8  per  cent.  ?  Ans.    $256.45. 

3.  Imported  200  boxes  of  figs,  weighing  60  Ib.  each  ;  draft 
was  reckoned  on  each  parcel  of  10  boxes,  and  tare  at  9  Ib.  per 
box.     What  was  the  duty  at  4  cents  per  Ib.  ?     Ans.    $404.80. 

4.  What  is  the  duty,  at  14  cents  per  pound,  on  6  barrels 
of  indigo,  weighing  450  Ib.  each,  the  legal  rate  of  tare  being  3 
percent.?  Ans.    $363.40+. 

5.  Required  the  duty,  at  2 1  cents  per  Ib.,  on  4  casks  of  shot, 
weighing  200  Ib.  each,  legal  tare  being  3  per  cent.    Ans.  $1 7.285. 

6.  Required  the  duty  on  3  pipes  of  port  wine,  gross  esti- 
mate as  follows:  —  first  pipe,  170  gallons;  second,  176  gal- 
lons ;  third,  180  gallons  ;  duty,  15  cents  per  gallon. 

Ans.    $77.322. 

7.  What  is  the  duty  on  250  dozen  bottles  of  porter,  at  5 
cents  per  bottle  ?  Ans.    $135. 

Give  rule  for  computing  ad  valorem  duties. 


SIMPLE  INTEREST. 


239 


SIMPLE  INTEREST. 

292.  Interest  is  a  sum  paid  for  the  use  of  money. 

293.  Principal  is  the  sum  for  the  use  of  which  interest 
is  paid. 

294.  Rate  per  cent,  per  annum  is  the  sum  per  cent,  paid 
for  the  use  of  $100  annually. 

NOTE.     The  rate  per  cent,  is  commonly  expressed  decimally,  as  hun- 
dredths  (231). 

290.    Amount  is  the  sum  of  the  principal  and  interest. 

296.  Simple  Interest  is  the  sum  paid  for  the  use  of  the 
principal  only,  during  the  whole  time  of  the  loan  or  credit. 

297.  Legal  Interest  is  the  rate  per  cent,  established  by 
law.     It  varies  in  different  States,  as  follows : 


Alabama, 8  per  cent. 


Arkansas, 6 

Connecticut, G 

Delaware, 6 

Dist.  of  Columbia,  ...  6 

Florida, 8 

Georgia, 8 

Illinois, 6 

Indiana, 6 

Iowa, 7 

Kentucky, 6 

Louisiana, 5 

Maine, 6 

Maryland, 6 

Massachusetts, 6 

Michigan, , ,  7 


Mississippi, . 


Missouri, 6 

New  Hampshire, 6 

New  Jersey, 6 

New  York, 7 

North  Carolina, 6 

Ohio, .6 

Pennsylvania, 6 

Rhode  Island, 6 

South  Carolina, 7 

Tennessee, 6 

Texas, 8 

!  United  States  (debts),  6 

|  Vermont, 6 

I  Virginia, 6 

i  Wisconsin, 7 


.  8  per  cent. 


NOTE.     When  the  rate  per  cent,  is  not  specified,  in  accounts,  notes, 
mortgages,  contracts,  &c.,  the  legal  rate  is  always  understood. 

29S.    Usury  is  illegal  interest,  or  a  greater  per  cent,  than 
the  legal  rate. 

CASE  i. 

299.    To  find  the  interest  on  any  sum,  at  any  rate 
per  cent.,  for  years  and  months. 

Define  interest.     Principal.     Rate  per  cent,  per  annum.     Amount, 
What  is  simple  interest  ?     Legal  interest  ?     Usury  ?     Case  I  ? 


240 


PERCENTAGE. 


In  percentage,  any  per  cent,  of  any  given  number  is  so 
many  hundredths  of  that  number  ;  but  in  interest,  any  rate  per 
cent,  is  confined  to  1  year,  and  the  per  cent,  to  be  obtained 
of  any  given  number  is  greater  than  the  rate  per  cent,  per 
annum  if  the  time  be  more  than  1  year,  and  less  than  the  rate 
per  cent,  per  annum  if  the  time  be  less  than  1  year.  Thus, 
the  interest  on  any  sum,  at  any  rate  per  cent.,  for  3  years  6 
months,  is  31  times  the  interest  on  the  same  sum  for  1  year ; 
and  the  interest  for  3  months  is  £  of  the  interest  for  1  year. 

1.  What  is  the  interest  on  $75.19  for  3  years  6  months,  at 
6  per  cent.  ? 


OPERATION. 

875.19 
.06 

$4.5114 


22557 
135342 


ANALYSIS.  The  interest  on  $75.19,  for  1  yr., 
at  6  per  cent,  is  .06  of  the  principal,  or  $4.51 14, 
and  the  interest  for  3  yr.  6  mo.  is  3fy  =.  31  times 
the  interest  for  1  yr.,  or  $4.5114  X  3£,  which  is 
$15.789  +i  the  Ans.  Hence,  the  following 


$15.7899,  Ans. 

RULE.  I.  Multiply  the  principal  by  the  rate  per  cent.,  and 
the  product  ivill  be  the  interest  for  1  year. 

II.  Multiply  this  product  by  the  time  in  years  and  fractions 
of  a  year,  and  the  result  will  be  the  required  interest. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  interest  of  $150  for  3  years,  at  4  per  cent.  ? 

Ans.    $18. 

3.  What  is  the  interest  of  $328  for  2  years,  at  7  pur  cent.  ? 

4.  What  is  the  interest  of  $125  for  1  year  6  months,  at  6 
percent.?  Ans.    Si  1.25. 

5.  What  is  the  interest  of  $200  for  3  years  10  months,  at 
7  per  cent  ?  Ans.'  $5:1(5(1  +  . 

6.  What  is  the  interest  of  $76.50  for  2  years  2  months,  at 
5  per  cent.  ? Ans.    $8.2<s7  -f- . 

Explain  the  difference  between  percentage  and  interest.  (iive 
analysis.  Rule. 


SIMPLE   INTEREST.  241 

7.  What  is  the  interest  of  $1270.25  for  11  months,  at  7 
per  cent.  ?  Ans.    $81.89  -f . 

8.  What  is  the  interest  of  $2569.75  for  4  years  6  months, 
at  6  per  cent.  ? 

9.  What  is  the  interest  of  $1500.60  for  2  years  4  months, 
at  6J-  per  cent.  ?  Ans.    $218.8375. 

10.  What  is  the  amount  of  $26.84  for  2  years  6  months,  at 
5  per  cent.  ?  Ans.    $30.195. 

11.  What  is  the  amount  of  $450  for  5  years,  at  7  per  cent.  ? 

12.  What  is  the  interest  of  $4562.09  for  3  years  3  months, 
at  3  per  cent.  ?  Ans     $444.80  -f- . 

13.  What  is  the  amount  of  $3050  for  4  years  8  months,  at 
5£  per  cent.  ?  Ans.    $3701.25. 

14.  What  is  the  interest  of  $5000  for  9  months,  at  8  per 
cent,  ?  Ans.    $300. 

15.  If  a  person  borrow  $375  at  7  per  cent,,  how  much  will 
be  due  the  lender  at  the  end  of  2  yr.  6  mo.  ? 

16.  What  is  the  interest  paid  on  a  loan  of  $1374.74,  at  6 
per  cent.,  made  January  1,  1856,  and  called  in  January  1, 
18GO?  Ans.    $329.937  +  . 

17.  If  a  note  of  $605.70  given  May  20, 1858,  on  interest  at 
8  percent.,  be  taken  up  May  20,  1861,  what  amount  will  then 
be  due  if  no  interest  has  been  paid?     Ans.   $751.068. 

CASE    II. 

3OO.    To  find  the  interest  on   any  sum,  for   any 
time,  at  any  rate  per  cent. 

The  analysis  of  our  rule  is  based  upon  the  following 

Obvious  Relations  between  Time  and  Interest. 

I.  The  interest  on  any  sum,  for  1  year,  at  1  per  cent.,  is 
.01  of  that  sum,  and  is  equal  to  the  principal  with  the  separatrix 
removed  two  places  to  the  left. 

II.  A  month  being  -fa  of  a  year,  -fa  of  the  interest  on  any 
sum  for  1  year  is  the  interest  for  1  month. 

What  is  Case  II  ?     Give  the  first  relation  between  time  and  interest. 
Second. 

U 


242  PERCENTAGE. 

III.  The  interest  on  any  sum  for  3  days  is  ^y  =  -^  =  .1 
of  the  interest  for  1  month,  and  any  number  of  days  may 
readily  be  reduced  to  tenths  of  a  month  by  dividing  by  3. 

IV.  The  interest  on  any  sum,  for  1  month,  multiplied  by 
any  given  time  expressed  in  months  and  tenths  of  a  month, 
will  produce  the  required  interest. 

1.   What  is  the  interest  on  $724.68  for  2  yr.  5  mo.  19  da., 
at  7  per  cent.  ? 

OPERATION.  ANALYSIS.      We  remove 

2yr.  5  mo.  19  da.  =  29.  6^  mo.      the  separatrix  in  the  given 
10  \  <fc7O/icQ  principal  two  places  to  the 

12)  $7-2468  left,  and  we  have  $7.2468, 

$.6039  the  interest  on  the  given  sum 

29.6£  for   1   year  at   1    per   cent. 

(300  L).     Dividing  this  by 
12,  we  have  $.6039,  the  inter- 
est  for  1  month,  at  1  per  cent. 
54351  (II.)       Multiplying    this 

12078  quotient  by  29.6^,  the  time 

expressed  in  months  and  deci- 
malg  of  a  month>  (m  IV  ^ 

we  have  $17.89557,  the  in- 


$125.26899,  Ans.  terest  on  the  given  sum  for 

the  given  time,  at  1  per  cent. 

(IV.).  And  multiplying  this  product  by  7  (7  times  1  per  cent.), 
we  have  $125.268  -f-,  the  interest  on  the  given  principal,  for  the 
given  time,  at  the  given  rate  per  cent.  Hence, 

RULE.  I.  Remove  the  separatrix  in  the  given  principal 
two  places  to  the  left  ;  the  result  will  be  the  interest  for  1  year, 
at  1  per  cent. 

II.  Divide  this  interest  byl2',  the  result  will  be  the  interest 
for  1  month,  at  1  per  cent. 

III.  Multiply  this  interest  by  the  given  time  expressed  in 
months  and  tenths  of  a  month  ;  the  result  will  be  the  interest 
for  the  given  time,  at  1  per  cent. 

IV.  Multiply  this  interest  by  the  given  rate  ;  the  product 
will  be  the  interest  required. 


Give  the  third.     Fourth.     Give  analysis.    Rule. 


SIMPLE  INTEREST.  243 

CONTRACTIONS.  After  removing  the  scparatrix  in  the  principal 
two  places  to  the  left,  the  result  may  be  regarded  either  as  the  in- 
terest on  the  given  principal  for  12  months  at  1  per  cent.,  or  for'l 
month  at  12  per  cent.  If  we  regard  it  as  for  1  month  at  12  per 
cent.,  and  if  the  given  rate  be  an  aliquot  part  of  12  per  cent.,  the 
interest  on  the  given  principal  for  1  month  may  readily  be  found  by 
taking  such  an  aliquot  part  of  the  interest  for  1  month  as  the  given 
rate  is  part  of  12  per  cent.  Thus, 

To  find  the  interest  for  1  month  at  6  per  cent.,  remove  the  sep- 
aratrix  two  places  to  the  left,  and  divide  by  2. 

To  find  it  at  3  per  cent,  proceed  as  before,  and  divide  by  4 ;  at  4 
per  cent.,  divide  by  3 ;  at  2  per  cent.,  divide  by  6,  &c. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  interest  of  $100  for  7  years  7  months,  at  6 
per  cent.  ?  Am.    $45.50. 

3.  What  is  the  amount  of  $47.50  for  4  years  1  month,  at  9 
percent?  Am.   $64.956  +  . 

4.  What  is  the  amount  of  $2000  for  3  months,  at  7  per 
cent?  Am.   $2035. 

5.  What  is  the  interest  of  $250  for  1  year  10  months  and 
15  days,  at  6  per  cent.  ?  Am.   $28.12^. 

6.  What  is  the  interest  of  $36.75  for  2  years  4  months  and 
12  days,  at  7  per  cent.  ?  Am.    $6.088  +  . 

7.  What  is  the  amount  of  $84  for  5  years  5  months  and  9 
days,  at  5  per  cent.  ? 

8.  What  is  the  interest  of  $51.10  for  10  months  and  3 
days,  at  4  per  cent.  ? 

9.  What  is  the  interest  of  $175.40  for  15  months  and  8 
days,  at  10  per  cent.  ?  Am.   $22.31  -f-  . 

10.  What  is  the  amount  of  $1500  for  6  months  and  24 
days,  at  7£  per  cent.  ?  Am.    $1563.75. 

11.  What  is  the  amount  of  $84.25  for  1  year  5  months 
and  10  days,  at  6£  per  cent.  ? 

12.  What  is  the  interest  of  $25  for  3  years  6  months  and 
20  days,  at  6  per  cent.  ?  Am.   $5.33^. 

13.  What  is  the  interest  of  $112.50  for  3  months  and  1 
day,  at  9 J-  per  cent.  ?  Am.    $2.70  +. 

What  contractions  are  given  ? 


244  PERCENTAGE. 

14   What  is  the  interest  of  $408  for  20  days,  at  6  per 
cent.?  Am.    $1.36. 

15.  What  is  the  interest  of  $500   for  22  days,  at  7  per 
cent.  ? 

16.  What  is  the  amount  of  $4500  for  10  days,  at  10  per 
cent.?  Ans.    $4512.50. 

17.  What  is  the  amount  of  $1000  for  1  month  5  days,  at 
6}  per  cent.  ?  Ans.    $1006.56£. 

18.  Find  the  interest  of  $973.68  for  7  months  9  days,  at 
4J-  per  cent. 

19.  If  I  borrow  $275  at  7  per  cent.,  how  much  will  I  owe 
at  the  end  of  4  months  25  days  ? 

20.  A  person  bought  a  piece  of  property  for  $2870,  and 
agreed  to  pay  for  it  in  1  year  and  6  months,  with  6J-  per  cent, 
interest;  what  amount  did  he  pay  ?  Ans.    $3149.825. 

21.  In  settling  with  a  merchant,  I  gave  my  note  for  $97.75, 
due  in  11  months,  at  5  per  cent.;  what  must  be  paid  when 
the  note  falls  due?  Ans.    $102.23  -f. 

22.  How  much  is  the  interest  on  a  note  of  $384.50  in  2 
years  8  months  and  4  days,  at  8  per  cent.  ? 

23.  What  is  the  interest  of  $97.86  from  May  17,  1850,  to 
December  19,  1857,  at  7  per  cent.  ?  Ans.    $51.98  +. 

24.  Find  the  interest  of  $35.61,  from  Nov.  11,  1857,  to 
Dec.  15,  1859,  at  6  per  cent.  Ans.    $4.474. 

25.  Required  the   interest  of  $50  from  Sept.  4,  1848,  to 
Jan.  1,  1860,  at  3£  per  cent. 

26.  Required  the  amount  of  $387.20,  from  Jan.  1  to  Oct. 
20,  1859,  at  7  per  cent.  Ans.    $408.957  +. 

27.  A  man,  owning  a  furnace,  sold  it  for  $6000 ;  the  terms 
were,  $2000  in  cash  on  delivery,  $3000  in  9  months,  and  the 
remainder  in  1    year   6  months,   with  7  per  cent,  interest; 
what  was  the  whole  amount  paid  ?  Ans.    $6262.50. 

28.  Wm.  Gallup  bought  bills  of  dry  goods  of  Geo.  Bliss 
&  Co.,  of  New  York,  as  follows,  viz.:   Jan.  10,  1858,  $350; 
April  15, 1858,  $150 ;  and  Sept.  20,  1858,  $550.50  ;  he  bought 
on  time,  paying  legal  interest ;    what  was  the  whole  amount 
of  his  indebtedness  Jan.  1,  1859  ?  Ans.   $1092.66  +. 


PARTIAL   PAYMENTS. 
PARTIAL  PAYMENTS   OR  IN 


OO1.    A  Partial  Payment  is  payment' 
bond,  or  other  obligation  ;  when  the  amount 
written  on  the  back  of  the  obligation,  it  becomes  a  receipt,  and 
is  called  an  Indorsement. 


$2000.  SPRINGFIELD,  MASS.,  Jan.  4,  1857. 

1.    For  value  received  I  promise  to  pay  James  Parish,  or 
order,  two  thousand  dollars,  one  year  after  date,  with  interest. 

GEORGE  JONES. 
On  this  note  were  indorsed  the  following  payments : 

Feb.  19,  1858, $400 

June  29,  1859, $1000 

Nov.  14,  1859, $520 

What  remained  due  Dec.  24,  1860? 

OPERATION. 

Principal  on  interest  from  Jan.  4,  1857,  $2000 

Interest  to  Feb.  19,  1858,  1  yr.  1  mo.  15  da., 135 


Amount, $2135 

Payment  Feb.  19,  1858, 400 


Remainder  for  a  new  principal, $1735 

Interest  from  Feb.  19,  1858,  to  June  29,  1859,  1  yr. 

4  mo.  10  da., 141.69 


Amount, $1876.69 

Payment  June  29,  1859, 1000. 

Remainder  for  a  new  principal, $876.69 

Interest  from  June  29,  1859,  to  Nov.  14,  1859,  4  mo. 

15  da., 19.725 

Amount, $896.415 

Payment  Nov.  14,  1859, 520. 

Remainder  for  a  new  principal, $376.415 

Interest  from  Nov.  14,  1859,  to  Dec.  24,  1860,  1  yr. 

1  mo.  10  da., 25.09 

Remains  due  Dec.  24,  1860, $401. 505 -f- 

What  is  meant  by  partial  payment  ?    By  an  indorsement  ? 

U* 


24G  PERCENTAGE. 

$475.50.  NEW  YORK,  May  1,  1855. 

2.  For  value  received,  we  jointly  and  severally  promise  to 
pay  Mason  &  Bro.,  or  order,  four  hundred  seventy-five  dol- 
lars fifty  cents,  nine  months  after  date,  with  interest. 

JONES,  SMITH  &  Co. 

The  following  indorsements  were  made  on  this  note : 

Dec.  25,  1855,  received, $50 

July  10,  1856,        «         15.75 

Sept.    1,  1857,        «        25.50 

June  14,  1858,        "         104 

How  much  was  due  April  15,  1859  ? 

OPERATION. 

Principal  on  interest  from  May  1,  1855, $475.50 

Interest  to  Dec.  25,  1855,  7  mo.  24  da., 21.63 


Amount, $497.13 

Payment  Dec.  25,  1855, 50. 

Remainder  for  a  new  principal, $447.14 

Interest  from  Dec.  25,  1855,  to  June  14,  1858,  2  yr. 

5  mo.  19  da., 77.29 

Amount, $524.42 

Payment  July  10,  1856,  less  than  interest") 

then  due, >  $15.75 

Payment  Sept.  1,  1857, J    25.50 

Their  sum  less  than  interest  then  due, . . .     $41.25 

Payment  June  14, 1858, 104. 

Their  sum  exceeds  the  interest  then  due, $145.25 


Remainder  for  a  new  principal, $379.17 

Interest  from  June  14,  1858,  to  April  15,  1859,  10  mo. 

1  da., 22.19 

Balance  due  April  15,  1859, $401.36  -f- 

These  examples  have  been  wrought  according  to  the  method 
prescribed  by  the  Supreme  Court  of  the  U.  S.,  and  are  suf- 
ficient to  illustrate  the  following 


PARTIAL   PAYMENTS.  247 

UNITED  STATES  RULE. 

I.  Find  the  amount  of  the  given  principal  to  the  time  of  the 
first  payment,  and  if  this  payment  exceed  the  interest  then  due 
subtract  it  from  the  amount  obtained,  and  treat  the  remainder 
as  a  new  principal. 

II.  But  if  the  interest  be  greater  than  any  payment,  cast  the 
interest  on  the  same  principal  to  a  time  when  the  sum  of  the 
payments  shall  equal  or  exceed  the  interest  due  ;  subtracting  the 
sum  of  the  payments  from  the  amount  of  the  principal,  the  re- 
mainder will  form  a  new  principal,  on  which  interest  is  to  be 
computed  as  before. 


SAN  FRANCISCO,  June  20,  1858. 

3.  Three  years  after  date  we  promise  to  pay  Ross  & 
Wade,  or  order,  five  hundred  fourteen  and  -f^  dollars,  for 
value  received,  with  10  per  cent,  interest.  WILDER  &  BRO. 

On  this  note  were  indorsed  the  following  payments :  Nov. 
12,  1858,  $105.50 ;  March  20,  1860,  $200  ;  July  10,  1860, 
$75.60.  How  much  remains  due  on  the  note  at  the  time  of 
its  maturity?  Ans.  $242.12-+. 


CHARLESTON,  May  7,  1859. 

4.    For  value  received,  I  promise  to  pay  George  Babcock 
three  thousand  dollars,  on  demand,  with  7  per  cent,  interest. 

JOHN  MAY. 

On  this  note  were  indorsed  the  following  payments  :  — 

Sept.  10,  1859,  received $25 

Jan.  1,  1860,  "       500 

Oct.  25,  1860,          "       75 

April  4,  1861,          "        1500 

How  much  was  due  Feb.  20,  1862  ?     Ans.    $1344.35  + . 


Give  the  United   States  Court  rule  for  computing  interest  where 
partial  payments  have  been  made. 


248  PERCENTAGE. 


NEW  ORLEANS,  Aug.  3,  1850. 


5.  One  year  after  date  I  promise  to  pay  George  Bailey,  or 
order,  nine  hundred  twelve  y7^  dollars,  with  5  per  cent,  in- 
terest, for  value  received.  JAMES  POWELL. 

The  note  was  not  paid  when  due,  but  was  settled  Sept.  15, 
1853,  one  payment  of  $250  having  been  made  Jan.  1,  1852, 
and  another  of  $316.75,  May  4,  1853.  How  much  was  due 
at  the  time  of  settlement  ?  Ans.  $467.53  . 


S184.56.  CINCINNATI,  April  2,  1860. 

6.  Four  months  after  date  I  promise  to  pay  J.  Ernst  & 
Co.  one  hundred  eighty-four  dollars  fifty-six  cents,  for  value 
received.  S.  ANDERSON. 

The  note  was  settled  Aug.  26,  1862,  one  payment  of  $50 
having  been  made  May  6,  1861.  How  much  was  due,  legal 
interest  being  6  per  cent.  ?  Ans.  $154.188  -f-  . 

NOTE.  A  note  is  on  interest  after  it  becomes  due,  if  it  contain  no 
mention  of  interest. 

7.  Mr.  B.  gave  a  mortgage  on  his  farm  for  $6000,  dated 
Oct.  1,  1851,  to  be  paid  in  6  years,  with  8  per  cent,  interest. 
Three  months  from  date  he  paid  $500  ;  Sept.  10,  1852,  $1126  ; 
March  31,  1854,  $2000  ;  and  Aug.  10,  1854,  $876.50.     How 
much  was  due  at  the  expiration  of  the  time  ?  Ans.  $3284.84  -f-  • 

«SO2.  The  United  States  rule  for  partial  payments  has 
been  adopted  by  nearly  all  the  States  of  the  Union  ;  the  only 
prominent  exceptions  are  Connecticut,  Vermont,  and  New 
Hampshire. 

CONNECTICUT  RULE. 

I.  Payments  made  one  year  or  more  from  the  time  the  in- 
terest commenced,  or  from  another  payment,  and  payments  less 
than  the  interest  due,  are  treated  according  to  the  United  States 
rule. 


Give  Connecticut  rule  for  partial  payments. 


PARTIAL  PAYMENTS.  249 

• 

II.  Payments  exceeding  the  interest  due,  and  made  within 
one  year  from  the  time  interest  commenced,  or  from  a  former 
payment,  shall  draw  interest  for  the  balance  of  the  year,  pro- 
vided  the  interval  does  not  extend  beyond  the  settlement,  and  the 
amount  must  be  subtracted  from  the  amount  of  the  principal  for 
one  year  ;  the  remainder  will  be  the  new  principal. 

III.  If  the  year  extend  beyond  the  settlement,  then  find  the 
amount  of  the  payment  to  the  day  of  settlement,  and  subtract  it 
from  the  amount  of  the  principal  to  that  day ;  the  remainder 
will  be  the  sum  due. 


WOODSTOCK,  CT.,  Jan.  1,  1858. 

1.  For  value  received,  I  promise  to  pay  Henry  Bowen,  or 
order,  four  hundred  sixty  dollars,  on  demand,  with  interest. 

JAMES  MARSHALL. 

On  this  note  are  indorsed  the  following  payments :  April 
16,  1858,  $148  ;  March  11,  1860,  $75  ;  Sept.  21,  1860,  $56. 
How  much  was  due  Dec.  11,  1860  ?  Ans.  $238.14+. 

8O"5,  A  note  containing  a  promise  to  pay  interest  an- 
nually is  not  considered  in  law  a  contract  for  any  thing  more 
than  simple  interest  on  the  principal.  For  partial  payments 
on  such  notes,  the  following  is  the 

VERMONT  EULE. 

I.  Find  the  amount  of  the  principal  from  the  time  interest 
commenced  to  the  time  of  settlement. 

II.  Find  the  amount  of  each  payment  from  the  time  it  was 
made  to  the  time  of  settlement. 

III.  Subtract  the  sum  of  the  amounts  of  the  payments  from 
the  amount  of  the  principal,  and  the  remainder  will  be  the 
sum  due. 


RUTLAND,  April  11,  1856. 


$600. 

1.    For  value  received,  I  promise  to  pay  Amos  Getting,  or 
order,  six  hundred  dollars  on  demand,  with  interest  annually. 

JOHN  BROWN. 

Give  the  Connecticut  rule  for  partial  payments.    The  Vermont  rule. 


250  PERCENTAGE. 

On  this  note  were  indorsed  the  following  payments  :  Aug. 
10,  1856,  $156;  Feb.  12,  1857,  $200;  June  1,  1858,  $185. 
What  was  due  Jan.  1,  1859  ?  Am.  $105.50+. 

SO4:.  In  New  Hampshire  interest  is  allowed  on  the  an- 
nual interest  if  not  paid  when  due,  in  the  nature  of  damages 
for  its  detention  ;  and  if  payments  are  made  before  one  year's 
interest  has  occurred,  interest  must  be  allowed  on  such  pay- 
ments for  the  balance  of  the  year.  Hence  the  following 

NEW  HAMPSHIRE  RULE. 

I.  Find  the  amount  of  the  principal  for  one  year,  and  de- 
duct from  it  the  amount  of  each  payment  of  that  year,  from  the 
time  it  was  made  up  to  the  end  of  the  year  ;  the  remainder  will 
be  a  new  principal,  with  which  proceed  as  before. 

II.  If  the  settlement  occur  less  than  a  year  from  the  last  an- 
nual term  of  interest,  make  the  last  term  of  interest  a  part  of  a 
year,  accordingly. 

®575-  KEENE,  N.  H.,  Aug.  4,  1858. 

1.  For  value  received,  I  promise  to  pay  George  Cooper,  or 
order,  five  hundred  seventy-five  dollars,  on  demand,  with  in- 
terest annually.  DAVID  GREENMAN. 

On  this  note  were  indorsed  the  following  payments :  Nov. 
4,  1858,  $64;  Dec.  13,  1859,  $48;  March  16,  1860,  $248; 
Sept.  28,  1860,  $60.  What  was  due  on  the  note  Nov.  4, 
1860?  Ans.  $215.3^. 

3O«>.  When  no  payment  whatever  is  made,  upon  a  note 
promising  annual  interest,  till  the  day  of  settlement,  in  New 
Hampshire  the  following  is  the 

COURT  RULE. 

Compute  separately  tJie  interest  on  the  principal  from  the 
time  the  note  is  given  to  the  time  of  settlement,  and  the  interest 
on  each  year's  interest  from  the  time  it  should  be  paid  to  the 
time  of  settlement.  Th.e  sum  of  the  interests  thus  obtained, 
added  to  the  principal,  will  be  the  sum  due. 

The  New  Hampshire  rule.    The  New  Hampshire  court  rule. 


PARTIAL   PAYMENTS.  251 


KEENE,  N.  H.,  Feb.  2f,  1855. 

1.  Three  years  after  date,  I  promise  to  pay  James  Clark, 
or  order,  five  hundred  dollars,  for  value  received,  with  interest 
annually  till  paid.  JOHN  S.  BRIGG-S. 

What  is  due  on  the  above  note,  Aug.  2, 1859  ?  Ans.  $649.40. 

PROBLEMS  IN  INTEREST. 

«SO6.  In  examples  of  interest  there  are  five  parts  involved, 
the  Principal,  the  Rate,  the  Time,  the"  Interest,  and  the 
Amount.  If  any  three  of  these  be  given,  the  others  may  be 
obtained. 

CASE    I. 

3O7.  The  time,  rate  per  cent.,  and  interest  being 
given,  to  find  the  principal. 

1.  What  principal  in  2  years,  at   6  per  cent.,  will  gain 
$31.80  interest  ? 

OPERATION.  ANALYSIS.,  Since $1, in 

S.I 2,  interest  of  $1  in  2 years  at  6 per  cent.          2  years,  at  6  per  cent.,  will 

$31.80  -I-  .12  =  $265,  Ans.          Sain  $-12  interest,  the  prin- 
cipal that  will  gain  $31.80, 

at  the  same  rate  and  time,  must  be  as  many  dollars  as  $.12  is  con- 
tained times  in  $31.80;  dividing,  we  obtain  $265,  the  required 
principal.  Hence, 

RULE.  Divide  the  given  interest  by  the  interest  of  $1  for 
the  given  time  and  rate,  and  the  quotient  will  be  the  principal. 

EXAMPLES    FOR    PRACTICE. 

2.  What  principal,  at  6  per  cent.,  will  gain  $28.12£  in  6 
years  3  months  ?  Ans.    $75. 

3.  What  sum,  put  at  interest  for  4  months  18  days,  at  4 
per  cent.,  will  gain  $9.20  ?  Ans.    $600. 

4.  What  sum  of  money,  invested  at  7  per  cent.,  will  pay 
me  an  annual  income  of  $1260  ?  Ans.    $18000. 

5.  What  sum  must  be  invested  in  real  estate,  yielding  10 
per  cent,  profit  in  rents,  to  produce  an  income  of  $3370  ? 
Ans.    $33700^ 

How  many  parts  are  considered  in  examples  in  interest  ?  What 
are  they  ?  What  is  Case  I  ?  Give  analysis.  Rule. 


252  PERCENTAGE. 

CASE   II. 

308.  The  time,  rate  per  cent.,  and  amount  being 
given,  to  find  the  principal. 

1.  What  principal  in   2  years    6   months,  at  7  per  cent, 
will  amount  to  $88.125? 

OPERATION.  ANALYSIS. 

$1.175  Amt.  of  $1  in  2  years  6  months,  at  7  per  cent.         Smce   $*'  m 

$88.125  -f-  1.175  =  $75,  Ans. 

months,  at  7 

per  cent.,  will  amount  to  $1.175,  the  principal  that  will  amount  to 
$88.125,  at  the  same  rate  and  time,  must  be  as  many  dollars  as 
$1.175  is  contained  times  in  $88.125 ;  dividing,  we  obtain  $75,  the 
required  principal.  Hence  the 

RULE.  Divide  the  given  amount  by  the  amount  of  $1  for 
the  given  time  and  rate,  Ofid  the  quotient  will  be  the  principal 
required. 

EXAMPLES    FOR    PRACTICE. 

2.  What  principal,  at  6  per  cent.,  will  amount  to  $655.20 
in  8  months  ?  Ans.    $630. 

3.  What  principal,  at  5  per  cent.,  will  amount  to  $106.855 
in  5  years  5  months  and  9  days  ?  Ans.    $84. 

4.  What  sum,  put  at  interest,  at  5£  per  cent,  for  8  years 
5  months,  will  amount  to  $1897.545  ?  Ans.    $1297.09+. 

5.  What  sum,  at  7  per  cent,  will  amount  to  $221.075  in  3 
years  4  months  ?  Ans.    $179.25. 

6.  What  is  the  interest  of  that  sum,  for  11  years  8  days,  at 
10£  per  cent,  which  will  at  the  given  rate  and  time  amount  to 
$857.54?  Ans.   $460.04. 

CASE    III. 

309.  The  principal,  time,  and  interest  being  given, 
to  find  the  rate  per  cent. 

1.  I  lent  $450  for  3  years,  and  received  for  interest  $67.50 ; 
what  was  the  rate  per  cent  ? 

Give  case  IE.    Analysis.    Rule.    Case  IH. 


PROBLEMS   IN   INTEREST.  253 

OPERATION.  ANALYSIS.    Since  at 

$  4.50  1  per  cent.  $450,  in  3 

3  years,  will  gain  $13.50 

interest,  the   rate   per 
$1 3.0 0,  int.  of  $450  for  3  years  at  1  per  cent.  ,  .  ,    ,, 

cent,  at  which  the  same 

$67.50  -f-  13.50  =  5  percent.,  Ans.         principal,  in  the  same 

time,  will  gain  $67.50, 

must  be  equal  to  the  number  of  times  $13.50  is  contained  in  $67.50; 
dividing,  we  obtain  5,  the  required  rate  per  cent.     Hence  the 

RULE.  Divide  the  given  interest  by  the  interest  on  the  prin- 
cipal for  the  given  time  at  1  per  cent.,  and  the  quotient  will  be 
the  rate  per  cent,  required. 

EXAMPLES    FOR    PRACTICE. 

2.  If  I   pay  $45    interest  for  the   use  of   $500   3  years, 
what  is  the  rate  per  cent.  ?  Ans.    3. 

3.  The  interest  of  $180  for  1   year   2  months   6  days  is 
$12.78  ;  what  is  the  rate  per  cent.  ?  Ans.    6. 

4.  A  man  invests    $2000   in  bank   stock,  and  receives  a 
semi-annual  dividend  of  $75 ;    what  is  the  rate  per  cent.  ? 

5.  At  what  per  cent,  must  $1000  be  loaned  for  3  years  3 
months  and  29  days,  to  gain  $183.18  ?  Ans.    5£. 

6.  A  man  builds  a  block  of  stores  at  a  cost  of  $2163:0,  and 
receives  for  them  an  annual  rent  of  $2596.80 ;  what  per  cent, 
does  he  receive  on  the  investment  ?  Ans.    12. 

CASE    IV. 

310.  Principal,  interest,  and  rate  per  cent,  being 
given,  to  find  the  time. 

1.   In  what  time  will  $360  gain  $86.40  interest,  at  6  per 

cent.  ? 

OPERATION.  ANALYSIS.     Sinee  in 

$  360  1  year  $360,  at  6  per 

.06  cent.,  will  gain  $21.60, 

the  number  of  years  in 
$21.60    Intereetof$360inlyearat6Percent.     wMch  ^   game   princi_ 

$86.40H-  21.60  =  4  years,  Ans.         pal,  at  the   same  rate, 

will  gain  $86.40,  will  be 

Analysis.     Rule.     Case  IV.     Analysis. 
V 


254  PERCENTAGE. 

as  many  as  $21.60  is  contained  times  in  $86.40 ;  dividing,  we  ob- 
tain 4  years,  the  required  time.     Hence  the 

RULE.  Divide  the  given  interest  by  the  interest  on  the  prin- 
cipal for  1  year,  and  the  quotient  will  be  the  time  required  in 
years  and  decimals. 

NOTE.  The  decimal  part  of  the  quotient,  if  any,  may  be  reduced  to 
months  and  days  (by  2O9). 

EXAMPLES    FOR    PRACTICE. 

2.  The  interest  of  $325  at  6  per  cent,  is  $58.50 ;  what  is 
the  time  ?  Ans.   3  years. 

3.  B  loaned  $1600  at  6  per  cent,  until  it  amounted  to 
$2000  ;  what  was  the  time  ?  Ans.    4  years  2  months. 

4.  How  long  must  $204  be  on  interest  at  7  per  cent.,  to 
amount  to  $217.09  ?  Ans.    11  months. 

5.  Engaging  in  business,  I  borrowed  $750  of  a  friend  at  6 
per  cent,  and  kept  it  until  it  amounted  to  $942  ;  how  long  did 
I  retain  it  ?  Ans.    4  years  3  months  6  days. 

6.  How  long  will  it  take  $200  to  double  itself  at  6  per  cent, 
simple  interest  ?  Ans.    1 6  years  8  months. 

7.  In  what  time  will  $675  double  itself  at  5  per  cent.  ? 

NOTE.  The  time  in  years  in  which  any  sum  will  double  itself  may 
be  found  by  dividing  100  by  the  rate  per  cent. 


COMPOUND  INTEREST. 

311.  Compound  Interest  is  interest  on  both  principal  and 
interest,  when  the  interest  is  not  paid  when  due. 

NOTE.  The  simple  interest  may  be  added  to  the  principal  annually, 
semi-annually,  or  quarterly,  as  the  parties  may  agree ;  but  the  taking 
of  compound  interest  is  not  legal. 

1.  What  is  the  compound  interest  of  $200,  for  3  years,  at 
6  per  cent.  ? 

Rule.  In  what  time  will  any  sum  double  itself  at  interest  ?  What 
is  compound  interest  ? 


COMPOUND   INTEREST.  255 

OPERATION. 

$200  Principal  for  1st  year. 

X  .06  z=       12          Interest  for  1st  year. 

$212          Principal  for  2d  year. 
$212  X  -06  =       12.72      Interest  for  2d  year. 

$224.72      Principal  for  3d  year. 
$224.72  X  -06  =       13.483    Interest  for  3d  year. 

$238.203    Amount  for  3  years. 
200.000    Given  principal. 

$38.203    Compound  interest. 

RULE.  I.  Find  the  amount  of  the  given  principal  at  the 
given  rate  for  one  year,  and  make  it  the  principal  for  the 
second  year. 

II.  Find  the  amount  of  this  new  principal,  and  make  it  the 
principal  for  the  third  year,  and  so  continue  to  do  for  the  given 
number  of  years. 

III.  Subtract  the  given  principal  from  the  last  amount,  and 
the  remainder  will  be  the  compound  interest. 

NOTES.  1.  "When  the  interest  is  payable  semi- annually  or  quar- 
terly, find  the  amount  of  the  given  principal  for  the  first  interval,  and 
make  it  the  principal  for  the  second  interval,  proceeding  in  all  respects 
as  when  the  interest  is  payable  yearly. 

2.  When  the  time  contains  years,  months,  and  days,  find  the  amount 
for  the  years,  upon  which  compute  the  interest  for  the  months  and 
days,  and  add  it  to  the  last  amount,  before  subtracting. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  compound  interest  of  $500  for  2  years  at  7 
per  cent.  ?  Ans.    $72.45. 

3.  What  is  the  amount  of  $312  for  3  years,  at  6  per  cent, 
compound  interest  ?  Ans.    $371.59-)-. 

4.  What  is  the  compound  interest  of  $250  for  2  years, 
payable  semi-annually,  at  6  per  cent.?  Ans.    $31.37-}-. 

5.  What  will  $450  amount  to  in  1  year,  at  7  per  cent,  com- 
pound interest,  payable  quarterly  ?  Ans.    $482.33. 

6.  What  is  the  compound  interest  of  $236  for  4  years  7 
months  and  6  days,  at  6  per  cent.  ?  Ans.    $72.66-}-. 

Explain  operation.     Give  rule. 


256 


PERCENTAGE. 


7.  What  is  the  amount  of  $700  for  3  years  9  months  and 
24  days,  at  7  per  cent,  compound  interest  ?  Ans.  $906.55+- 

A  more  expeditious  method  of  computing  compound  interest 
than  the  preceding,  is  by  means  of  the  following 

TABLE, 

Showing  the  amount  of$l,  or  £1,  at  3,  4,  5,  6,  and  7  per  cent.,  compound 
interest,  for  any  number  of  years,  from  1  to  20. 


Yrs. 

3  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

1 

1.030,000 

1.040,000 

1.050,000 

1.060,000 

1.07,000 

2 

1.060,900 

1.081,600 

1.102,500 

1.123,600 

1.14,490 

3 

1.092,727 

1.124,864 

1.157,625 

1.191,016 

1.22,504 

4 

1.125,509 

1.169,859 

1.215,506 

1.262,477 

1.31,079 

5 

1.159,274 

1.216,653 

1.276,282 

1.338,226 

1.40,255 

6 

1.194,052 

1.265,319 

1.340,096 

1.418,519 

1.50,073 

7 

1.229,874 

1.315,932 

1.407,100 

1.503,630 

1.60,578 

8 

1.266,770 

1.368,569 

1.477,455 

1.593,848 

1.71,818 

9 

1.304,773 

1.423,312 

1.551,328 

1.689,479 

1.83,845 

10 

1.343,916 

1.480,244 

1.628,895 

1.790,848 

1.96,715 

11 

1.384,234 

1.539,454 

1.710,339 

1.898,299 

2.10,485 

12 

1.425,761 

1.601,032 

1.795,856 

2.012,196 

2.25,219 

13 

1.468,534 

1.665,074 

1.885,649 

2.132,928 

2.40,984 

14 

1.512.590 

1.731,676 

1.979,932 

2.260,904 

2/57,853 

15 

1.557,967 

1.800,944 

2.078,928 

2.396,558 

2.75,903 

16 

1.604,706 

1.872,981 

2.182,875 

2.540,352 

2.95,216 

17 

1.652,848 

1.947,900 

2.292,018 

2.692,773 

3.15,881 

18 

1.702,433 

2.025,817 

2.406,619 

2.854,339 

3.37,293 

19 

1.753,506 

2.106,849 

2.526,950 

3.025,600 

3.61,652 

20 

1.806,111 

2.191,12312.653,298 

3.207,135  3.86,968 

8.  What  is  the  amount  of  $800  for  6  years,  at  7  per  cent. 

OPERATION. 

From  the  table     $1.50073        Amount  of  $1  for  the  time, 
800    Principal. 

$1200.58400,     Ans. 

9.  What  is  the  compound  interest  of  $120  for  15  years,  at 
5  per  cent.?  Ans.    $129.47+. 

Of  what  use  is  the  table  in  computing  compound  interest  ? 


DISCOUNT.  257 

10.  What  is  the  amount   of  $.10    for  20  years,  at  7  per 
cent.?  Ans.    $.38696. 

DISCOUNT. 

81'2.  Discount  is  an  abatement  or  allowance  made  for 
the  payment  of  a  debt  before  it  is  due. 

313.  The  Present  Worth  of  a  debt,  payable  at  a  future 
time  without  interest,  is  such  a  sum  as,  being  put  at  legal  in- 
terest, will  amount  to  the  given  debt  when  it  becomes  due. 

1.  A  owes  B  $321,  payable  in  1  year;  what  is  the  pres- 
ent worth  of  the  debt,  the  use  of  money  being  worth  7  per 
cent.  ? 

OPERATION.  ANALYSIS.  The 

Am'tof     $1,  1.07)  $321  ($300,  Present  value,    amount  of  $1   for 

321  !  Year  is   $1-07; 

therefore  the  pres- 

$321   Given  sum  or  debt.  ent  WQrth  of  eyery 

300  Present  ™rth.  $1.07  of  the  given 

$21  Discount.  debt    is    *l*    and 

the  present  worth 

of  $321  will  be  as  many  dollars  as  $1.07  is  contained  times  in  $321. 
$321  -^  1.07  =  $300,  Ans.     Hence  the  following 

RULE.  I.  Divide  the  given  sum  or  debt  by  the  amount  of 
$1  for  the  given  rate  and  time,  and  the  quotient  will  be  the  pres- 
ent worth  of  the  debt.  9 

11.  Subtract  the  present  worth  from  the  given  sum  or  debt) 
and  the  remainder  will  be  the  discount. 

NOTE.  The  terms  present  worth,  discount,  and  debt,  are  equivalent 
to  principal,  interest,  and  amount.  Hence,  when  the  time,  rate  per 
cent.,  and  amount  are  given,  the  principal  may  be  found  by  (3O8)  » 
and  the  interest  by  subtracting  the  principal  from  the  amount. 

EXAMPLES    FOB    PRACTICE. 

2.  What  is  the  present  worth  of  $180,  payable  in  3  years 
4  months,  discounting  at  6  per  cent.  ?  Ans.    $150. 

Define  discount.     Present  worth.     Give  analysis.     Rule. 


258  PERCENTAGE. 

3.  What  is  the  present  worth  of  a  note  for  $1315.389,  due 
in  2  years  6  months,  at  7  per  cent.?  Ans.    $1119.48. 

4.  What  is  the  present  worth  of  a  note  for  $866.038,  due 
in  3  years  6  months  and  6  days,  when  money  is  worth  8  per 
cent.?     What  the  discount  ?  Ans.    $190.15-)-,  discount. 

5.  What  is  the  present  worth  of  a  debt  for  $1005,  on  which 
$475  is  to  be  paid  in  10  months,  and  the  remainder  in  1  year  3 
months,  the  rate  of  interest  being  6  per  cent.  ? 

NOTE.  When  payments  are  to  be  made  at  different  times  without 
interest,  find  the  present  worth  of  each  payment  separately,  and  take 
their  sum. 

Ans.   $945.40+. 

6.  I  hold  a  note  against  C  for  $529.925,  due  Sept.  1, 1859 ; 
what  must  I  discount  for  the  payment  of  it  to-day,  Feb.  7, 
1859,  money  being  worth  6  per 'cent.  ?  Ans.    $17.425. 

7.  A  man  was  offered  $3675   in   cash  for  his  house,  or 
$4235  in  3  years,  without  interest;    he  accepted  the  latter 
offer ;  how  much  did  he  lose,  money  being  worth  7  per  cent.  ? 

Ans.    $175. 

8.  A  man,  having  a  span  of  horses  for  sale,  offered  them 
for  $480  cash  in  hand,  or  a  note  of  $550  due  in  1  year  8 
months,  without  interest ;  the  buyer  accepted  the  latter  offer  ; 
did  the  seller  gain  or  lose  by  his  offer,  and  how  much,  interest 
being  6  per  cent.  ?  Ans.    Seller  lost  $20. 

9.  What  must  be  discounted  for  the   present  payment  of  a 
debt  of  $2637.72,  of  which  $517.50  is  to  be  paid  in  6  months, 
$793.75  in  10  months,  and  the  remainder  in  1  year  6  months, 
the  use  of  money  being  worth  7  per  cent.  ?      Ans.  $187.29  +. 

10.  What  is  the  difference  between  the  interest  and  discount 
of  $130,  due  10  months  hence,  at  10  per  cent,  ?     Ans.  $.83£. 

PROMISCUOUS    EXAMPLES    IN    PERCENTAGE. 

1.  A  merchant  bought  sugar  in  New  York  at  0]  rents  per 
pound  ;  (lie.  wastage  by  transportation  and  retailing  was  5  per 
cent.,  and  the  interest  on  the  first  cost  to  the  time  of  sale  was 
2  per  cent. ;  how  much  must  he  ask  per  pound  to  gain  25  per 
cent.  ?  Ans.  8  cents. 


PROMISCUOUS   EXAMPLES.  259 

2.  A  person  purchased  2  lots  of  land  for  $200  each,  and     ,  ^ 
sold  one  at  40  per  cent,  more  than  cost,  and  the  other  at  20 
per  cent,  less  ;  how  much  did  he  gain  ?  Ans.    $40. 

3.  Sold  goods  to  the  amount  of  $425,  on  6  months'  credit, 
which  was  £25  more  than  the  goods  cost;  what  was  the  true 
profit,  money  being  worth  6  per  cent.  ?        Ans.    $12.62  -(-. 

4.  Bought  cotton  cloth  at  13  cents  a  yard,  on  8  months' 
credit,  and  sold  it  the  same  day  at  12  cents  cask;  how  much    * 
did  I  gain  or  lose  per  cent.,  money  being  worth  6  per  cent.  ? 

Ans.    Lost  4  per  cent. 

5.  A  farmer  sold  a  pair  of  horses  for  $150  each;  on  one 
he  gained  25  per  cent.,  on  the  other  he  lost  25  per  cent. ;  did 
he  gain  or  lose  on  both,  and  how  much  ?       Ans.    Lost  $20. 

6.  A  man  invested  §  of  all  he  was  worth  in  the  coal  trade, 
and  at  the  end  of  2  years  8  months  sold  out  his  entire  interest 
for  $3100,  which  was  a  yearly  gain  of    9  per  cent,  on  the 
money  invested ;  how  much  was  he  worth  when  he  commenced 
trade?  Ans.    $3750. 

7.  In  how  many  years  will  a  man,  paying  interest  at  7  per 
cent,  on  a  debt  for  land,  pay  the  face  of  the  debt  in  interest  ? 

Ans.    1 4f  years. 

8.  Two  persons  engaged  in  trade ;  A  furnished  f  of  the 
capital,  and  B  f;  and  at  the  end  of  3 'years  4  months  they 
found  they  had  made  a  clear  profit  of  $5000,  which  was  124- 
per  cent,  per  annum  on  the  money  Invested  ;   how  much  cap- 
ital did  each  furnish  ?  Ans.   A,  $7500  ;  B,  $4500. 

9.  Bought  $500  worth  of  dry  goods,  and  $800  worth  of 
groceries ;  on  the  dry  goods  I  lost  20  per  cent.,  but  on  the 
groceries  I  gained  15  per  cent. ;    did  I  gain  or  lose  on  the 
whole  investment,  and  how  much  ?  Ans.   Gained  $20. 

10.  What  amount  of  accounts  must  an  attorney  collect,  in 
order  to  pay  over  $1100,  and  retain  8^-  per  cent,  for  collect- 
ing? Ans.    $1200. 

11.  A  merchant  sold  goods  to  the  amount  of  $667,  to  be 
paid  in  8  months ;   the  same  goods  cost  him  $600  one  year 
previous  to  the  sale  of  them ;  money  being  worth  6  per  cent., 
what  was  his  true  gain  ?  •      Ans.    $5.346  -(-. 


260  PERCENTAGE. 

12.  A  nurseryman  sold  trees  at    $18   per   hundred,  and 
cleared  ^  of  his  receipts ;   what  per  cent,  profit  did  he  make  ? 

Ans.    50  per  cent. 

13.  If  |  of  an  article  be  sold  for  what  f-  of  it  cost,  what  is 
the  gain  per  cent.  ?  .  Ans.    40|. 

14.  A  lumber  merchant  sells  a  lot  of  lumber,  which  he  has 
had  on  hand  6  months,  on  10  months'  credit,  at  an  advance  of 
30  per  cent,  on  the  first  cost ;  if  he  is  paying  5  per  cent,  inter- 
est on  capital,  what  are  his  profits  per  cent.  ?      Ans.    21 J-J-. 

15.  A  person,  owning  |  of  a  piece  of  property,  sold  20  per 
cent,  of  his  share ;  what  part  did  he  then  own  ?       Ans.   ^. 

16.  A  speculator,  having  money  in  the  bank,  drew  60  per 
cent,  of  it,  and  expended  40  per  cent,  of  50  per  cent,  of  this  for 
728  bushels  of  wheat,  at  $1.50  per  bushel ;   how  much  was 
left  in  the  bank  ?  Ans.    $3640. 

17.  I  wish  to  line  the  carpet  of  a  room,  that  is  6  yards  long 
and  5  yards  wide,  with  duck  £  yard  wide ;  how  many  yards  of 
lining  must  I  purchase,  if  it  will  shrink  4  per  cent,  in  length, 
and  5  per  cent,  in  width  ?  Ans.   43f  f . 

18.  A's  money  is  28  per  cent,  more  than  B's ;  how  many 
per  cent,  is  B's  less  than  A's  ?  Ans.   21£. 

19.  A  capitalist  invested  -|  of  his  money  in  railroad  stock, 
which  depreciated  5  per  cent,  in  value ;  the  remaining  f  he  in- 
vested in  bank  stock,  which,  at  the  end  of  1  year,  had  gained 
$1200,  which  was  12  per  cent,  of  the  investment ;  what  was  the 
whole  amount  of  his  capital,  and  what  was  his  entire  loss  or 
gain  ?  Ans.    $25000,  capital ;  $450,  gain. 

20.  C's  money  is  to  D's  as  2  to  3 ;  if  %  of  C's  money  be 
put  at  interest  for  3  years  9  months,  at  10  per  cent.,  it  will 
amount  to  $1933.25 ;  how  much  money  has  each  ? 

Ans.    C,  $2812  ;  D,  $4218. 

BANKING. 
314.    A  Bank  is  a  corporation  chartered  by  law  for  the 

purpose  of  receiving  and  loaning  money,  and   furnishing  a 
paper  circulation. 

What  is  a  bank? 


BANKING.  261 

315.  A  Promissory  Note  is  a  written  or  printed  engage- 
ment to  pay  a  certain  sum,  either  on  demand  or  at  a  specified 
time. 

31G.  Bank  Notes,  or  Bank  Bills,  are  the  notes  made 
and  issued  by  banks  to  circulate  as  money.  They  are  payable 
in  specie  at  the  banks. 

317.  The  Face  of  a  note  is  the  sum  made  payable 'by 
the  note. 

318.  Days  of  Grace  are  the  three  days  usually  allowed 
by  law  for  the  payment  of  a  note  after  the  expiration  of  the 
time  specified  in  the  note. 

319.  The  Maturity  of  a  note  is  the  expiration  of  the 
days  of  grace  ;  a  note  is  due  at  maturity. 

320.  Notes  may  contain  a  promise  of  interest,  which  will 
be  reckoned  from  the  date  of  the  note,  unless  some  other  time 
be  specified. 

The  transaction  of  borrowing  money  at  banks  is  conducted 
in  accordance  with  the  following  custom :  the  borrower  pre- 
sents a  note,  either  made  or  indorsed  by  himself,  payable  at 
a  specified  time,  and  receives  for  it  a  sum  equal  to  the  face, 
less  the  interest  for  the  time  the  note  has  to  run.  The  amount 
thus  withheld  by  the  bank  is  in  consideration  of  advancing 
money  on  the  note  prior  to  its  maturity. 

321.  Bank  Discount  is  an  allowance  made  to  a  bank  for 
the  payment  of  a  note  before  it  becomes  due. 

322.  The  Proceeds  of  a  note  is  the  sum  received  for  it 
when  discounted,  and  is  equal  to  the  face  of  the  note  less  the 
discount. 

CASE    I. 

323.  Given  the  face  of  a  note  to  find  the  proceeds. 
The  law  of  custom  at  banks  makes  the  discount  of  a  note 


Define  a  promissory  note.  Bank  notes.  The  face  of  a  note.  Days 
of  grace.  The  maturity  of  a  note.  Explain  the  process  of  discounting 
a  note  at  a  bank.  Define  bank  discount.  The  proceeds  of  a  note. 
What  is  Case  I  ? 


262  PERCENTAGE. 

equal  to  the  simple  interest  at  the  legal  rate  for  the  time  spe- 
cified in  the  note.     Hence  the 

RULE.  I.  Compute  the  interest  on  the  face  of  the  note  for 
three  days  more  than  the  specified  time  ;  the  result  will  be  the 
discount. 

II.  Subtract  the  discount  from  the  face  of  the  note,  and  the 
remainder  will  be  the  proceeds. 

EXAMPLES    FOR   PRACTICE. 

1.  "What  is  the  discount,  and  what  the  proceeds,  of  a  note 
for  $450,  at  60  days,  discounted  at  a  bank  at  6  per  cent.  ? 

Ans.    Discount,  $4.725  ;  proceeds,  $445.275. 

2.  What  are  the  proceeds  of  a  note  for  $368,  at  90  days, 
discounted  at  the  Bank  of  New  York  ?      Ans.    $361.345  -[-. 

3.  What  shall  I  receive  on  my  note  for  $475.50,  at  60 
days,  if  discounted  at  the  Crescent  City  Bank,  New  Orleans  ? 

Ans.    $471.33+. 

4.  What  are  the  proceeds  of  a  note  for  $10000,  at  90  days, 
discounted  at  the  Philadelphia  Bank  ?  Ans.    $9845. 

5.  Paid,  in  cash,  $240  for  a  lot  of  merchandise.     Sold  it 
the  same  day,  receiving  a  note  for  $250  at  60  days,  which  I 
got  discounted  at  the  Hartford  Bank.     What  did  I  make  by 
this  speculation  ?  Ans.    $7.37£. 

6.  A  note  for  $360.76,  drawn  at  90  days,  is  discounted  at 
the  Vermont  Bank.     Find  the  proceeds.      Ans.   $355.1  68 -f-. 

7.  Wishing  to  borrow   $530  of  a  western  bank  which  is 
discounting  paper  at  8  per  cent.,  I  give  my  note  for  $536.75, 
payable  in  60  days.     How  much  do  I  need  to  make  up  the 
required  amount  ?  Ans.    $.7(>-ir>. 

NOTES.  1.  To  indicate  the  maturity  of  a  note  or  draft,  a- vortical 
line  (  |  )  is  used,  with  the  day  at  which  the  note  is  nominally  due  on 
the  left,  and  the  date  of  maturity  on  the  right ;  thus,  Jan.  '  \  10 . 

2.  When  a  note  is  on  interest,  payable  at  a  future  specified  time,  the 
amount  is  the  face  of  the  note,  or  the  sum  made  payable,  and  must  be 
made  the  basis  of  discount. 

Give  rule. 


BANKING. 


Find  the  maturity,  term  of  discount,  and  proceeds  of  the 


following  notes  :  — 


$500.  BOSTON,  Jan.  4,  1859. 

8.  Three  months  after  date,  I  promise  to  pay  to  the  order  of 
John  Brown  &  Co.  five  hundred  dollars,  at  the  Suffolk  Bank, 
value  received.  JAMES  BARKER. 

Discounted  March  2.  (  Due,  April  4|7. 

Am.   <  Term  of  discount,  36  da. 

(  Proceeds,  $497. 

$750.  ST.  Louis,  June  12,  1859. 

9.  Six  months  after  date,  I  promise  to  pay  Thomas  Lee,  or 
order,  seven  hundred  fifty  dollars,   with  interest,  value  re- 
ceived. BYRON  QUINBY. 

Discounted  at  a  broker's,  Nov.  15,  at  10  per  cent. 

(Due,  Dec."|1B. 

Ans.    I  Term  of  discount,  30  da. 

(Proceeds,    $766.434+. 

CASE    II. 

324.  Given  the  proceeds  of  a  note,  to  find  the 
face. 

1.  I  wish  to  borrow  $400  at  a  bank.  For  what  sum  must 
I  draw  my  note,  payable  in  60  days,  so  that  when  discounted 
at  6  per  cent.  I  shall  receive  the  desired  amount  ? 

OPERATION.  ANALYSIS.     $400  is  the 

$1.0000  proceeds  of  a  certain  note, 

.0105  —  disc,  on  $1  for  63  da.  the   face    of  which    we    are 

required  to  find.    We  first 

$  .9895  =  proceeds  of  $1.  obtain  the  proceeds  of  $1 

$400  H-  -9895  =  $404.244  =         by  the  last  case,  and  then 

face  of  the  required  note.  divide   the   given   proceeds, 

$400,  by  this  sum ;  for,  as  many  times  as  the  proceeds  of  $1  is  con- 
tained in  the  given  proceeds,  so  many  dollars  must  be  the  face  of 
the  required  note.  Hence  the 

Give  Case  II.     Analysis. 


264  PERCENTAGE. 

RULE.  Divide  the  proceeds  by  the  proceeds  of  $1  for  the 
time  and  rate  mentioned,  and  the  quotient  will  be  the  face  of 
the  note. 

EXAMPLES    FOR   PRACTICE. 

2.  What  is  the  face  of  a  note  at  60  days,  which  yields 
$680  when  discounted  at  a  New  Haven  bank  ? 

Ans.    $687.215+. 

3.  What  is  the  face  of  a  note  at  90  days,  of  which  the  pro- 
ceeds are  $1000  when  discounted  at  a  Louisiana  bank  ? 

Ans.    $1013.085+. 

4.  Wishing  to  borrow  $500  at  a  bank,  for  what  sura  must 
my  note  be  drawn,  at  30  days,  to  obtain  the  required  amount, 
discount  being  at  7  per  cent.  ?  Ans.    $503.22  +. 

5.  James  Hopkins  buys  merchandise  of  me  in  New  York, 
at  cash   price,  to  the  amount  of  $1256.     Not  having  money, 
he  gives  his  note  in  payment,  drawn  at  6  months.     What 
must  be  the  face  of  the  note  ?  Ans.   $1302.341  +. 


EXCHANGE. 

Exchange  is  a  method  of  remitting  money  from  one 
place  to  another,  or  of  making  payments  by  written  orders. 

•!'3G.  A  Bill  of  Exchange  is  a  written  request  or  order 
upon  one  person  to  pay  a  certain  sum  to  another  person,  or  to 
his  order,  at  a  specified  time. 

•SS7.  A  Sight  Draft  or  Bill  is  one  requiring  payment  to 
be  made  "  nt  sight,"  which  means,  at  the  time  of  its  presenta- 
tion to  the  person  ordered  to  pay.  In  other  bills,  the  time 
specified  is  usually  a  certain  number  of  days  "after  sight." 

There  are  always  three  parties,  and  usually  four,  to  a  trans- 
action in  exchange  : 

328.  The  Drawer  or  Maker  is  the  person  who  signs  the 
order  or  bill. 


the  rule.    Define  exchange.    A  bill  of  exchange.    A  si^lit  draft. 
The  drawer. 


EXCHANGE.  265 

329.  The  Drawee  is  the  person  to  whom  the  order  is 
addressed. 

330.  The  Payee  is  the  person  to  whom  the  money  is  or- 
dered to  be  paid. 

331.  The  Buyer  or  Remitter  is  the  person  who  purchases 
the  bill.     He  may  be  himself  the  payee,  or  the  bill   may  be 
drawn  in  favor  of  any  other  person. 

332.  The  Indorsement  of  a  bill  is  the  writing  upon  its 
back,  by  which  the  payee  relinquishes  his  title,  and  transfers 
the  payment  to  another.     The  payee  may  indorse  in  blank  by 
writing  his  name  only,  which  makes  the  bill  payable  to  the 
bearer,  and  consequently  transferable  like  a  bank  note  ;  or  he 
may  accompany  his  signature  by  a  special  order  to  pay  to 
another  person,  who  in  his  turn  may  transfer  the  title  in  like 
manner.      Indorsers   become   separately  responsible  for   the 
amount  of  the  bill,  in  case  the  drawee  fails  to  make  payment. 
A  bill  made  payable  to  the  bearer  is  transferable  without  in- 
dorsement. 

333.  The  Acceptance  of  a  bill  is  the  promise  which  the 
drawee  makes  when  the  bill  is  presented  to  him  to  pay  it  at 
maturity  ;  this  obligation  is  usually  acknowledged  by  writing 
the  word  "  Accepted,"  with  his  signature,  across  the  face  of 
the  bill. 


Three  days  of  grace  are  usually  allowed  for  the  payment  of 
a  bill  of  exchange  after  the  time  specified  has  expired.  But  in  New 
York  State  no  grace  is  allowed  on  sight  drafts. 

From  these  definitions,  the  use  of  a  bill  of  exchange  in  mon- 
etary transactions  is  readily  perceived.  If  a  man  wishes  to 
make  a  remittance  to  a  creditor,  agent,  or  any  other  person 
residing  at  a  distance,  instead  of  transporting  specie,  which  is 
attended  with  expense  and  risk,  or  sending  bank  notes,  which 
are  liable  to  be  uncurrent  at  a  distance  from  the  banks  that 
issued  them,  he  remits  a  bill  of  exchange,  purchased  at  a  bank 
or  elsewhere,  and  made  payable  to  the  proper  person  in  or 

The  drawee.  The  payee.  The  buyer.  An  indorsement.  An 
acceptance.  What  of  grace  on  bills  of  exchange  ? 

W 


266  PERCENTAGE. 

near  the  place  where  he  resides.  Thus  a  man  by  paying 
Boston  funds  in  Boston,  may  put  New  York  funds  into  the 
hands  of  his  New  York  agent. 

334.  The  Course  of  Exchange  is  the  variation  of  the 
cost  of  sight  bills  from  their  par  value,  as  affected  by  the  rela- 
tive conditions  of  trade  and  commercial  credit  at  the  two  places 
between  which  exchange  is  made.     It  may  be  either  at  a  pre- 
mium or  discount,  and  is  rated  at  a  certain  per  cent,  on  the 
face  of  the  bill.     Bills  payable  a  specified  time  after  sight  are 
subject  to  discount,  like  notes  of  hand,  for  the  term  of  credit 
given.     Hence  their  value  in  the  money  market  is  affected  by 
both  "the  course  of  exchange  and  the  discount  for  time. 

335.  Foreign  Exchange  relates  to  remittances  made  be- 
tween different  countries. 

336.  Domestic    or   Inland  Exchange   relates  to  remit- 
tances made  between  different  places  in  the  same  country. 

An  inland  bill  of  exchange  is  commonly  called  a  Draft. 
In  this  work  we  shall  treat  only  of  Inland  Exchange. 

CASE    I. 

337.  To  find  the  cost  of  a  draft. 


$500.  SYRACUSE,  May  7,  1859. 

1.   At  sight,  pay  to  James  Clark,  or  order,  five  hundred 
dollars,  value  received,  and  charge  the  same  to  our  account. 
To  M.  SMITH  &  Co. 

Messrs.  BROWN  &  FOSTER,  ) 

Baltimore. 

What  is  the  cost  of  the  above  draft,  the  rate  of  exchange 
being  1£  per  cent,  premium? 

OPERATION.  ANALYSIS.    Since  ex- 

$500  X  1 .01 5  =  $507.50,  Ans.         chan?e  is  at  ^  ST, Cent; 

premium,  each  dollar  ot 

the  draft  will  cost  $1.015  ;  and  to  find  the  whole  cost  of  the  draft, 

How  is  exchange  conducted  ?     Explain  course  of  exchange.     For- 
eign exchange.     Inland  exchange.     Define  a  draft.     What  is  Case  I 
Give  analysis. 


EXCHANGE.  267 

we  multiply  its  face,  $500,  by  1.015,  and  obtain  $507.50,  the  re- 
quired Ans. 

$480.  BOSTON,  June  12,  1859. 

2.  Thirty  days  after  sight,  pay  to  John  Otis,  or  bearer,  four 
hundred  eighty  dollars,  value  received,  and  charge  the  same 
to  account  of  AMOS  TRENCHARD. 

To  JOHN  STILES  &  Co., 
New  York. 

What  is  the  cost  of  the  above  draft,  exchange  being  at  a 
premium  of  3  per  cent.  ? 

OPERATION.  ANALYSIS.       Since 

$1.0000  time    is     allowed,    the 

.0055  z=z  discount  for  33  days.  draft   must   suffer   dis- 

"TTr  count  in  the  sale.     The 

$    .9945=:  proceeds  of  $1.  ,. 

discount  of  $1,  at  the 

.Uo       z=  rate  of  exchange.  i        i       L     •     T»  p 
legal  rate  in  Boston,  for 

$1.0245=  cost  of  $1  of  the  draft.  the   specified  time,  al- 

$480  X  1.0245  =  $491.76,  Ans.         lo™g  grace,  is  $.0055, 

which,  subtracted  from 

$1,  gives  $.9945,  the  cost  of  $1  of  the  dr.aft,  provided  sight  ex- 
change were  at  par  ;  but  sight  exchange  being  at  premium,  we  add 
the  rate,  .03,  to  .9945,  and  obtain  $1.0245,  the  actual  cost  of  $1. 
Then,  multiplying  $480  by  1.0245,  we  obtain  $491.76,  the  Ans. 
From  these  examples  we  derive  the  following 

RULE.  I.  For  sight  drafts.  —  Multiply  the  face  of  the  draft 
by  1  plus  the  rate  when  exchange  is  at  a  premium,  and  by 
1  minus  the  rate  when  exchange  is  at  a  discount. 

II.  For  drafts  payable  after  sight.  —  Find  the  proceeds  of  $1 
at  bank  discount  for  the  specified  time,  at  the  legal  rate  where 
the  draft  is  purchased;  then  add  the  rate  of  exchange  when 
at  a  premium,  or  subtract  it  when  at  a  discount,  and  multiply 
the  face  of  the  draft  by  this  result. 

EXAMPLES    FOR    PRACTICE. 

3.  A  merchant  in   Cincinnati  wishes  to  remit  $1000  by 

Give  analysis.     Rule  I;  IE. 


268  PERCENTAGE. 

\ 

draft  to  his  agent  in  New  York ;  what  will  the  bill  cost,  ex- 
change being  at  3  per  cent,  premium  ?  Am.  $1030. 

4.  What  will  be  the  cost  in  Rochester  of  a  draft  on  Albany 
for  $400,  payable  at  sight,  exchange  being  at  f  per  cent,  pre- 
mium? Ans.    $403. 

5.  A  merchant  in  St.  Louis  orders  goods  from  New  York, 
to  the  amount  of  $530,  which  amount  he  remits  by  draft,  ex- 
change being  at   2f  per  cent,  premium.     If  he  pays  $20  for 
transportation,  what  will  the  goods  cost  him  in  St.  Louis  ? 

Ans.   $564.575. 

6.  What  will  be  the  cost,  in  Detroit,  of  a  draft  on  Boston 
for  $800,  payable  60  days  after  sight,  exchange  being  at  a  pre- 
mium of  2  per  cent.  ?  Ans.    $806.20. 

7.  A  man  in  Philadelphia  purchased  a  draft  on  Chicago  for 
$420,  payable  30  days  after  sight ;   what  did  it  cost  him,  the 
rate  of  exchange  being  1^  per  cent,  discount?    Ans.  $411.30. 

8.  A  merchant  in  Portland  receives  from  his  agent  320 
barrels  of  flour,  purchased  in  Chicago  at  $10  per  barrel ;  in 
payment  for  which  he  remits  a  draft  on  Chicago,  at  2£  per 
cent,  discount.      The  transportation  of  his  flour  cost  $312. 
What  must  he  sell  it  for  per  barrel  to  gain  $400  ?     Ans.  $12. 

CASE    II. 

338.  To  find  the  face  of  a  draft  which  a  given  sum 
will  purchase. 

1.  A  man  in  Indiana  paid  $369.72  for  a  draft  on  Boston, 
drawn  at  30  days ;  what  was  the  face  of  the  draft,  exchange 
being  at  3  J  per  cent,  premium  ? 

OPERATION.  ANALYSIS.     We  find, 

$369.72  —  1.027  =  $360,  Ans.        b^  Case  T' that  a  draft 

for  $1  will  cost  $1.027; 

hence  the  draft  that  will  cost  $369.72  must  he  for  as  many  dollars  as 
1.027  is  contained  times  in  $369.72;  dividing,  we  obtain  813(30,  the 
Ans.  From  this  example  and  analysis  we  derive  the  following 

What  is  Case  II  ?     (Jive  analysis. 


EQUATION  OF  PAYMENTS.  269 

RULE.  Divide  the  given  cost  by  the  cost  of  a  draft  for  $1, 
at  the  given  rate  of  exchange  ;  the  quotient  will  be  the  face  of 
the  required  draft. 

EXAMPLES    FOR    PRACTICE. 

2.  What  draft  may  be  purchased  for  $243. GO,  exchange 
being  at  1£  per  cent,  premium?  Ans.    $240. 

3.  What  draft  may  be  purchased  for  $79.20,  exchange  be- 
ing at  1  per  cent,  discount  ?  Ans.    $80. 

4.  An  agent  in   Pittsburg   holding  $282.66,  due  his  em- 
ployer in  New  Haven,  is    directed  to  make  the   remittance 
by  draft,  drawn  at  60  days.     What  will  be  the  face  of  the 
draft,  exchange  being  at  2  per  cent,  premium  ?       Ans.    $280. 

5.  An  emigrant  from  Bangor  takes  $240  in  bank  bills  to 
St.  Paul,  Min.,  and  there  pays  £  per  cent,  brokerage  in  ex- 
change for  current  money.     What  would  he  have  saved  by 
purchasing  in  Bangor  a  draft  on  St.  Paul,  drawn  at  30  days, 
exchange  being  at  1£  per  cent,  discount  ?  Ans.    $5.60. 

6.  A  Philadelphia  manufacturer  is  informed  by  his  agent  in 
Buffalo  that  $3600  is  due  him  on  the  sale  of  some  property. 
He  instructs  the  agent  to  remit  by  a  draft  payable  in  60  days 
after  sight,  exchange  being  at  f  per  cent,  premium.     The  agent, 
by  mistake,  remits  a  sight  draft,  which,  when  received  in  Phila- 
delphia, is  accepted,  and  paid  after  the  expiration  of  the  three 
days  of  grace.     If  the  manufacturer  immediately  puts  this 
money  at  interest  at  the  legal  rate,  will  he  gain  or  lose  by  the 
blunder  of  his  agent  ?  Ans.   He  will  lose  $8.24-|-. 

EQUATION  OF  PAYMENTS. 

339.  Equation  of  Payments  is  the  process  of  finding  the 
mean  or  equitable  time  of  payment  of  several  sums,  due  at 
different  times  without  interest. 

340.  The  Term  of  Credit  is  the  time  to  elapse  before  a 
debt  becomes  due. 

Bole.    Define  equation  of  payments.     Term  of  credit. 
W* 


270  EQUATION   OF  PAYMENTS. 

5141 .  The  Average  Terra  of  Credit  is  the  time  to  elapse 
before  several  debts,  due  at  different  times,  may  all  be  paid 
at  once,  without  loss  to  debtor  or  creditor. 

342.  The  Equated  Time  is  the  date  at  which  the  several 
debts  may  be  canceled  by  one  payment. 

CASE    I. 

343.  When  all  the  terms  of  credit  begin  at  the 
same  date. 

1.  On  the  first  day  of  January  I  find  that  I  owe  Mr.  Smith 
8  dollars,  to  be  paid  in  5  months,  10  dollars  to  be  paid  in  2 
months,  and  12  dollars  to  be  paid  in  10  months ;  at  what  time 
may  I  pay  the  whole  amount  ? 

OPERATION. 
$  8  X    5  =    40 
10  X    2  =    20 
12  X  10  =  120 

30  180  -7-  30  =  6  mo.,  average  time  of  credit. 

Jan.  1.  -f-  6  mo.  =  July  1,  equated  time  of  payment. 
ANALYSIS.  The  whole  amount  to  be  paid,  as  seen  above,  is  $30 : 
and  we  are  to  find  how  long  it  shall  be  withheld,  or  what  term  of 
credit  it  shall  have,  as  an  equivalent  for  the  various  terms  of  credit 
on  the  different  items.  Now,  the  value  of  credit  on  any  sum  is  meas- 
ured by  the  product  of  the  money  and  time.  And  we  say,  the  credit 
on  $8  for  5  mo.  =  the  credit  on  $40  for  1  mo.,  because  8  X  5  =  40 
X  1.  In  the  same  manner,  we  have,  the  credit  on  $10  for  2  mo.z=: 
the  credit  on  $20  for  1  mo. ;  and  the  credit  on  $12  for  10mo.:= 
the  credit  on  $120  for  1  mo.  Hence,  by  addition,  the  value  of  the 
several  terms  of  credit  on  their  respective  sums  equals  a  credit  of  1 
month  on  $180;  and  this  equals  a  credit  of  6  months  on  $30,  be- 
cause 

30X6z=180X  1. 

RULE.  I.  Multiply  each  payment  by  its  term  of  credit,  and 
divide  the  sum  of  tie  products  In/  the  sum  of  the  payments  ;  the 
quotient  will  be  the  average  term  of  credit. 

Average  term  of  credit.  Equated  time.  Give  Case  I.  Analysis. 
Bule. 


AVERAGING   CREDITS.  271 

II.  Add  the  average  term  of  credit  to  the  date  at  which  all 
the  credits  begin,  and  the  result  will  be  the  equated  time  of 
payment. 

NOTES.  1.  The  periods  of  time  used  as  multipliers  must  all  be  of 
the  same  denomination,  and  the  quotient  will  be  of  the  same  denomi- 
nation as  the  terms  of  credit ;  if  these  be  months,  and  there  be  a  re- 
mainder after  the  division,  continue  the  division  to  days  by  reduction, 
always  taking  the  nearest  unit  in  the  last  result. 

2.  The  several  rules  in  equation  of  payments  are  based  upon  the 
principle  of  bank  discount ;  for  they  imply  that  the  discount  of  a  sum 
paid  before  it  is  due  equals  the  interest  of  the  same  amount  paid  after 
it  is  due. 

EXAMPLES    FOR    PRACTICE. 

2.  On  the  25th  of  September  a  trader  bought  merchandise, 
as  follows :  $700  on  20  days'  credit ;  $400  on  30  days'  credit ; 
$700  on  40  days'  credit :  what  was  the  average  term  of  credit, 
and  what  the  equated  time  of  payment  ? 

A        (  Average  credit,  30  days. 

(.  Equated  time  of  payment,  Oct.  25. 

3.  On  July  1  a  merchant  gave  notes,  as  follows :  the  first 
for  $250,  due  in  4  months ;   the  second  for  $750,  due  in  2 
months ;  the  third  for  $500,  due  in  7  months :  at  what  time 
may  they  all  be  paid  in  one  sum  ?  Ans.   Nov.  1. 

4.  A  farmer  bought  a  cow,  and  agreed  to  pay  $1  on  Mon- 
day, $2  on  Tuesday,  $3  on  Wednesday,  and  so  on  for  a  week ; 
desirous  afterward  to  avoid  the  Sunday  payment,  he  offered  to 
pay  the  whole  at  one  time :  on  what  day  of  the  week  would 
this  payment  come  ?  Ans.    Friday. 

5.  Jan.  1,  I  find  myself  indebted  to  John  Kennedy  in  sums 
as  follows :  $650  due  in  4  months ;  $725  due  in  8  mouths ;  and 
$500  due  in  12  months :  at  what  date  may  I  settle  by  giving 
my  note  on  interest  for  the  whole  amount  ?    Ans.  Aug.  20. 

CASE    II. 

344.  When  the  terms  of  credit  begin  at  different 
dates,  and  the  account  has  only  one  side. 

345.  An  Account  is  the  statement  or  record  of  mercantile 
transactions  in  business  form. 

Give  Case  II.     Define  an  account. 


272 


EQUATION   OF   PAYMENTS. 


3<1G.  The  Items  of  an  account  may  be  sums  due  at  the 
date  of  the  transaction,  or  on  credit  for  a  specified  time. 

An  account  may  have  both  a  debit  and  a  credit  side,  the 
former  marked  Dr.,  the  latter  Cr.  Suppose  A  and  B  have 
dealings  in  which  there  is  an  interchange  of  money  or  prop- 
erty ;  A  keeps  the  account,  heading  it  with  B's  name ;  the  Dr. 
side  of  the  account  shows  what  B  has  received  from  A ;  the 
Cr.  side  shows  what  he  has  parted  with  to  A. 

347.  The  Balance  of  account  is  the  difference  of  the  two 
sides,  and  may  be  in  favor  of  either  party. 

If,  in  the  transactions,  one  party  has  received  nothing  from 
the  other,  the  balance  is  simply  the  whole  amount,  and  the 
account  has  but  one  side.  Bills  of  purchase  are  of  this  class. 

NOTE.  Book  accounts  bear  interest  after  the  expiration  of  the  term 
of  credit,  and  notes  after  they  become  due. 

348.  To  Average  an  Account  is  to  find  the  mean  or 
equitable  time  of  payment  of  the  balance. 

349.  A  Focal  Date  is  a  date  to  which  all  the  others  are 
compared  in  averaging  an  account. 

1.    When  does  the  amount  of  the  following  bill  become  due, 
by  averaging  ? 
J.  C.  SMITH, 

1859.  To  C.  E.  BORDEN,      Dr. 

June    1.     To  Cash, $450 

"     12.      "    Mdse.  on  4  mos., 500 

Aug.  16.      "    Mdse., 250 


FIRST   OPERATION. 


SECOND   OPERATION. 


Due. 

da. 

Items. 

Prod. 

June  1 

0 

450 

Oct.  12 

133 

500 

66500 

Aug.  16 

70 

250 

19000 

1200 

85500 

Due. 

da. 

Items. 

Prod. 

June  1 

133 

450 

59850 

Oct.  12 

0 

500 

Aug.  16 

57 

250 

14250 

1200 

74100 

85500 -f- 1200  =  71  da. 
A  ^  71  da.  after  June  1, 
Ans'\orAug.  11. 


74100-^1200  =  62  da. 

<  62  da.  before  Oct.  12, 
Ans'\  or  Aug.  11. 


Define  items.     Balance.    To  average  an  account.    A  focal  date. 


AVERAGING   ACCOUNTS.  273 

ANALYSIS.  By  reference  to  the  example,  it  will  be  seen  that  the 
items  are  due  June  1,  Oct.  12,  and  Aug.  16,  as  shown  in  the  two 
operations.  In  the  first  operation  we  use  the  earliest  date,  June  1, 
as  a  focal  date,  and  find  the  difference  in  days  between  this  date  and 
each  of  the  others,  regard  being  had  to  the  number  of  days  in  cal- 
endar months.  From  June  1  to  Oct.  12  is  133  da. ;  from  June  1  to 
Aug.  16  is  76  da.  Hence  the  first  item  has  no  credit  from  June  1, 
the  second  item  has  133  days'  credit  from  June  1,  and  the  third 
item  has  76  days'  credit  from  June  1,  as  appears  in  the  column 
marked  da.  After  this  we  proceed  precisely  as  in  Case  I,  and  find 
the  average  credit, 71  da.,  and  the  equated  time,  Aug.  11. 

In  the  second  operation,  the  latest  date,  Oct.  12,  is  taken  for  a 
focal  date ;  the  work  is  explained  thus :  Suppose  the  account  to  be 
settled  Oct.  12.  At  that  time  the  first  item  has  been  due  133  days, 
and  must  therefore  draw  interest  for  this  time.  But  interest  on 
$450  for  133  days  =  the  interest  on  $59850  for  1  da.  The  second 
item  draws  no  interest,  because  it  falls  due  Oct.  12.  The  third  item 
must  draw  interest  57  days.  But  interest  on  $250  for  57  days  = 
the  interest  on  $14250  for  1  day.  Taking  the  sum  of  the  products, 
•  we  find  the  whole  amount  of  interest  due  on  the  account,  at  Oct.  12, 
equals  the  interest  on  $74100  for  1  day;  and  this,  by  division,  is 
found  to  be  equal  to  the  interest  on  $1200  for  62  days,  which  time 
is  the  average  term  of  interest.  Hence  the  account  would  be  settled 
Oct.  12,  by  paying  $1200  with  interest  on  the  same  for  62  days.  This 
shows  that  1200  has  been  due  62  days  ;  that  is,  it  falls  due  Aug.  11, 
without  interest.  Hence  the  following 

RULE.  I.  Find  the  time  at  which  each  item  becomes  due, 
by  adding  to  the  date  of  each  transaction  the  term  of  credit,  if 
any  be  specified,  and  write  these  dates  in  a  column. 

II.  Assume  either  the  earliest  or  the  latest  date  for  a  focal 
date,  and  find  the  difference  in  days  between  the  focal  date  and 
each  of  the  other  dates,  and  write  the  results  in  a  second  column. 

III.  Write  the  items  of  the  account  in  a  third  column,  and 
multiply  each  sum  by  the  corresponding  number  of  days  in  the 
preceding  column,  writing  the  products  in  a  final  column. 

IV.  Divide  the  sum  of  the  products  by  the  sum  of  the  items. 
The  quotient  will  be  the  average  term  of  credit  when  the 

Give  analysis.    Rule. 


274  EQUATION   OP   PAYMENTS. 

earliest  date  is  the  focal  date,  or  the  average  term  of  interest 
when  the  latest  date  is  the  focal  date ;  in  either  case  always 
reckon  from  the  focal  date  toward  the  other  dates,  to  find  the 
equated  time  of  payment. 

EXAMPLES  FOR  PRACTICE. 

2.  JOHN  BROWN, 

1859o  To  JAMES  GREIGG,      Dr. 

Jan.     1.     To      50  yds.  Broadcloth,  fa)  $3.00,  .  .  .  $150 
"      16.      «   2000    "     Calico,          "       .10, 0  .  .     200 
Feb.     4.      «        75    «     Carpeting,    "      1.33£,  .  .     100 
March  3.      «      400    "     Oil  Cloth,     "       .40,...     160 
If  James  Greigg  wishes  to  settle  the  above  bill  by  giving 
his  note,  from  what  date  shall  the  note  draw  interest  ? 

Ans.   Jan.  28. 

3.  ABRAM  RUSSEL, 

1859,  To  WYNKOOP  &  BRO.,    Dr. 

March  1.     To  Cash,  .  .  ,  .  0  .  «  e  „ .  . > ,  $300 

April    4.      "    Mdse.,  .  0  .  ...  0  . 0  <,„  0  0  «...„.    240 

June  18.      "        "       on  2  mo.,  0  . ,  „  0  0  0  „  c  *  » .    100 
Aug.     8.      "    Cash, .  .  0  *  o ,  0  „  „  „  .  o  .  o  .  .  .  . . .    400 

"What  is  the  equated  time  of  payment  of  the  above  account  ? 

Ans.   May  26. 

4.  JOHN  OTIS, 

1858.  To  JAMES  LADD,       Dr. 

June  1.       To  500  bu.  Wheat,  fa)  $1.20, ..,,..  $600 

•"    12.        "    200   "        "        «     1.50, 300 

"   15.        "    640   "        "        "     1.30, .....  o    832 

"   25.        <fc    760   "        "       «     1.00,,  .<•*.    760 

"   30.        "    500   «        «       "     1.50, ,  . .  ,  ..    750 

When  is  the  whole   amount  of  the   above   bill  due,  per 

average  ?  Ans.   June  18. 

5.  My  expenditures  in  building  a  Louse,  in  the  year  1856, 
were  as  follows :  Jim.  1C,  $536.78  ;  Feb.  20,  $425.36 ;  March  4, 
$259.25 ;  April  24,  $786.36.     If  at  the  last  date  I  agree  to 


AVERAGING   ACCOUNTS. 


275 


sell  the  house  for  exactly  what  it  cost,  with  reference  to  interest 
on  the  money  expended,  and  take  the  purchaser's  note  for  the 
amount,  what  shall  be  the  face  of  the  note,  and  what  its 
date  ?  Am  f  Face,  $2007.75. 

"  \  Date,  March  8,  1856. 
6.   THOMAS  WHITING, 
1859.  To  ISRAEL  PALMER,     Dr. 

Jan.     L     To    60  bbls.  Flour,   (3>  $7.00, $420 

"     28.      «     90  bu.     Wheat,  «      1.50,  ...    .    135 
Mar.  15.       "   300  bbls.  Flour,     "      6.00,  . .     .  .  1800 
If  credit  of  3  months  be  given  to  each  item,  when  will  the 
above  account  become  due  ?  Ans.  May  27. 


CASE  in. 

When  the  terms  of  credit  begin  at  different 
times,  and  the  account  has  both  a  debt  and  a  credit 
side. 

1.    Average  the  following  account. 

DAVID  WARE. 
Dr.  Cr. 


185 
June 

3. 
1 

To  Mdse 

400 

00 

1851 
July 

3. 
4 

By  Mdse 

200 

Of) 

« 

Ifi 

"   Draft  3  mo 

800 

00 

Au0* 

<>0 

"   Cash 

150 

00 

Oct. 

20 

"   Cash,  

250 

00 

Sept. 

20 

500 

00 

Dr. 


OPERATION. 


Cr. 


Foral  ) 
date.  \ 

Due 

da. 

Items. 

Prod. 

Due 

July     4 
Aug.  20 
Sept.  20  1 

da. 

Items. 

Prod. 

June    1 
Sept.  19 
Oct.  20 

141 
31 
0 

400 

800 
250 

56400 
24800 

108 

61 

1    30 

200 
150 
500 

21600 
9150 
15000 

Balances. 

1450 
850 

81200 
45750 

850     45750 

600 

35450 

35450  -7-  600  =.  59  da.,  average  term  of  interest. 
Oct.  20  —  59  da.  =  Aug.  22,  balance  due. 


What  is  Case  III  r     Explain  operation. 


276 


EQUATION   OP   PAYMENTS. 


ANALYSIS.  In  the  above  operation  we  have  written  the  dates, 
showing  when  the  items  become  due  on  either  side  of  the  ac- 
count, adding  3  days'  grace  to  the  time  allowed  to  the  draft.  The 
latest  date,  Oct.  20,  is  assumed  as  the  focal  date  for  both  sides,  and 
the  two  columns  marked  da.  show  the  difference  in  days  between 
each  date  and  the  focal  date.  The  products  are  obtained  as  in  the 
last  case,  and  a  balance  is  struck  between  the  items  charged  and  the 
products.  These  balances,  being  on  the  Dr.  side,  show  that  David 
Ware,  on  the  day  of  the  focal  date,  Oct.  20,  owes  $600  with  interest 
on  $35450  for  1  day.  By  division,  this  interest  is  found  to  be  equal 
to  the  interest  on  $600  for  59  days.  The  balance,  $600,  therefore, 
has  been  due  59  days.  Reckoning  back  from  Oct.  12,  we  find  the 
date  when  the  balance  fell  due,  Aug.  22.  Hence  the  following 

RULE.  I.  Find  the  time  when  each  item  of  the  account  is 
due  ;  and  write  the  dates,  in  two  columns,  on  the  sides  of  the 
account  to  which  they  respectively  belong. 

II.  Use  either  the  earliest  or  the  latest  of  these  dates  as  the 
focal  date  for  both  sides,  and  Jind  the  products  as  in  the  last 
case. 

III.  Divide  the  balance  of  the  products  by  the  balance  of  the 
account;  the  quotient  will  be  the  interval  of  time,  which  must 
be  reckoned  from  the  focal  date  TOWARD  the  other  dates  when 
both  balances  are  on  the  same  side  of  the  account,  but  FROM 
the  other  dates  when  the  balances  are  on  opposite  sides  of  the 
account. 

2.  What  is  the  balance  of  the  following  account,  and  when 
is  it  due  ? 


JOHN  WILSON. 


Dr. 


Cr. 


185 

9. 

185 

9. 

Jan. 

1 

To  Mdse.  . 

448 

00 

Jan. 

20 

By  Am't  bro't  forward 

560 

00 

Feb. 

4 

"  Cash.  . 

364 

00 

Feb. 

16 

264 

00 

20 

«      « 

232 

00 

a 

25 

"  Cash  

900 

00 

A 


Balance,  $680. 
Due  March  13. 


3.  If  the  following  account  be  settled  by  giving  a  note,  what 
shall  be  the  face  of  the  note,  and  what  its  dale? 


Give  analysis.    Rule. 


RATIO. 


277 


Dr. 


ISAAC  FOSTER. 


Cr. 


185£ 

185* 

J. 

-J  an 

1 

To  Mdse.  on  3  mo. 

145 

86 

May 

11 

By  Cash  

11 

00 

" 

12 

«       a        «   5    « 

37 

48 

July 

12 

a       « 

15 

00 

June 

Aug. 

3 
4 

.<         «          «     2     « 

12 
66 

25 

48 

Oct. 

12 

82 

00 

Ans. 


$154.07,  face  of  note. 
Mar.  26,  1858,  date. 


RATIO. 

•l«>  t .  Ratio  is  the  comparison  with  each  other  of  two  num- 
bers of  the  same  kind.  It  is/of  two  kinds  —  arithmetical  and 
geometrical. 

352.  Arithmetical  Ratio  is  the   difference   of  the   two 
numbers. 

353.  Geometrical  Ratio  is  the  quotient  of  one  number 
divided  by  the  other. 

354.  When  we  use  the  word  ratio  alone,  it  implies  geo- 
metrical ratio,  and  is  expressed  by  the  quotient  arising  from 
dividing  one  number  by  the  other.     Thus,  the  ratio  of  4  to  8 
is  2,  of  10  to  5  is  £,  &c. 

355.  Ratio  is  indicated  in  two  ways. 

1st.  By  placing  two  points  between  the  numbers  compared, 
writing  the  divisor  before  and  the  dividend  after  the  points. 
Thus,  the  ratio  of  5  to  7  is  written  5  :  7  ;  the  ratio  of  9  to 
4  is  written  9  :  4. 

2d.  In  the  form  of  a  fraction ;  thus,  the  ratio  of  9  to  3  is  J ; 
the  ratio  of  4  to  6  is  f . 

356.  The  Terms  are  the  two  numbers  compared. 

357.  The  Antecedent  is  the  first  term. 

358.  The  Consequent  is  the  second  term. 

359.  No  comparison  of  two  numbers  can  be  fully  ex- 
plained but  by  instituting  another  comparison  ;  thus,  the  com- 


NOTE.  It  is  thought  best  to  omit  the  questions  at  the  bottom  of  the  pa^es.  in  the 
remaining  pfH  of  this  work,  leaving  the  teacher  to  use  such  as  may  be  deemed  ap- 
propriate. 


278  RATIO. 

parison  or  relation  of  4  to  8  cannot  be  fully  expressed  by  2, 
nor  of  8  to  4  by  J.  If  the  question  were  asked,  what  relation 
4  bears  to  8,  or  8  to  4,  in  respect  to  magnitude,  the  answer  2, 
or  £,  would  not  be  complete  nor  correct.  But  if  we  make 
unity  the  standard  of  comparison,  and  use  it  as  one  of  the 
terms  in  illustrating  the  relation  of  the  two  numbers,  and 
say  that  the  ratio  or  relation  of  4  to  8  is  the  same  as  1  to  2, 
or  the  ratio  of  8  to  4  is  the  same  as  1  to  %,  unity  in  both  cases 
being  the  standard  of  comparison,  then  the  whole  meaning  is 
conveyed. 

36O.  A  Direct  Ratio  arises  from  diivding  the  consequent 
by  the  antecedent. 

261.  An  Inverse  or  Reciprocal  Ratio  is  obtained  by  di- 
viding the  antecedent  by  the  consequent.  Thus,  the  direct 
ratio  of  5  to  15  is  -^5-  =  3 ;  and  the  inverse  ratio  of  5  to  15  is 

*==* 

362.  A  Simple  Ratio  consists  of  a  single  couplet ;   as 
3  :  12. 

363.  A  Compound  Ratio  is  the  product  of  two  or  more 
simple   ratios.     Thus,  the   compound  ratio  formed  from  the 
simple  ratios  of  3  :  6  and  8  :  2  is  f  X  f  =  3  X  8  :  6  X  2  = 

it=*. 

364:.  In  comparing  numbers  with  each  other,  they  must 
be  of  the  same  kind,  and  of  the  same  denomination. 

365.  The  ratio  of  two  fractions  is  obtained  by  dividing 
the  second  by  the  first ;  or  by  reducing  them  to  a  common  de- 
nominator, when  they  are  to  each  other  as  their  numerators. 
Thus,  the  ratio  of  -fr  :  f  is  f  -^  &  =  f  f  —  2,  which  is  the 
same  as  the  ratio  of  the  numerator  3  to  the  numerator  6  of 
the  equivalent  fractions  -ft-  and  TV 

Since  the  antecedent  is  a  divisor  and  the  consequent  a  divi- 
dend, any  change  in  either  or  both  terms  will  be  governed  by 
the  general  principles  of  division,  (87. )  We  have  only  to 
substitute  the  terms  antecedent,  consequent,  and  ratio,  for  divi- 
sor, dividend,  and  quotient,  and  these  principles  become 


RATIO.  279 


GENERAL  PRINCIPLES   OF  RATIO. 

PRIN.  I.  Multiplying  the  consequent  multiplies  the  ratio ; 
dividing  the  consequent  divides  the  ratio. 

PRIN.  II.  Multiplying  the  antecedent  divides  the  ratio  ;  di- 
viding the  antecedent  multiplies  the  ratio. 

PRIN.  III.  Multiplying  or  dividing  both  antecedent  and  con- 
sequent by  the  same  number  does  not  alter  the  ratio. 

These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

A  change  in  the  consequent  produces  a  LIKE  change  in  the 
ratio  ;  but  a  change  in  the  antecedent  produces  an  OPPOSITE 
change  in  the  ratio. 

36O.  Since  the  ratio  of  two  numbers  is  equal  to  the  con- 
sequent divided  by  the  antecedent,  it  follows,  that 

1.  The  antecedent  is  equal  to  the  consequent  divided  by 
the  ratio ;  and  that, 

2.  The  consequent  is  equal  to  the  antecedent  multiplied  by 
the  ratio. 

EXAMPLES    FOR   PRACTICE. 

1.  What  part  of  9  is  3? 

f  =  i;  or,  9  :  3  as  1  :  £,  that  is,  9  has  the  same  ratio  to  3  that  1 
b#s  to  \. 

2.  What  part  of  20  is  5  ?  Ans.   f 

3.  What  part  of  36  is  4?  Ans.   £. 

4.  What  part  of  7  is  49  ?  Ans.   7  times. 

5.  What  is  the  ratio  of  16  to  88  ?  Ans.    5%. 

6.  What  is  the  ratio  of  6  to  8£?  Ans.   |£. 

7.  What  is  the  ratio  of  6J-  to  78  ?  Ans.    12. 

8.  What  is  the  ratio  of  16  to  66  ?  Ans.    4|. 

9.  What  is  the  ratio  of  £  to  f  ?  Ans.    f . 

10.  What  is  the  ratio  of  f  to  T4F?  Ans.    f. 

11.  What  is  the  ratio  of  3^  to  16|?  Ans.    5. 

12.  What  is  the  ratio  of  3  gal.  to  2  qt.  1  pt.  ?      Ans.   ^. 


280  PROPORTION. 

13.  What  is  the  ratio  of  6.3  s  to  8  s.  6  d.  ?         Ans.    Iff. 

14.  What  is  the  ratio  of  5.6  to  .56  ?  Ans.    TV 

15.  What  is  the  ratio  of  19  Ibs.  5  oz.  8  pwts.  to  25  Ibs.  11 
oz.  4  pwts.  ?  Ans.   !-£. 

16.  What  is  the  inverse  ratio  of  12  to  16?  Ans.   }. 

17.  What  is  the  inverse  ratio  of  f  to  J  ?  Ans.   T9¥. 

18.  What  is  the  inverse  ratio  of  5J  to  17£?         Ans.    ^. 

19.  If  the  consequent  be  16  and  the  ratio  2f,  what  is  the 
antecedent  ?  Ans.   7. 

20.  If  the  antecedent  be  14.5  and  the  ratio  3,  what  is  the 
consequent  ?  Ans.    43.5. 

21.  If  the  consequent  be  J  and  the  ratio  £,  what  is  the  an- 
tecedent? Ans.    1£. 

22.  If  the  antecedent  be  f  and  the  ratio  £,  what  is  the 
consequent  ?  Ans.   •£?• 

PROPORTION. 

367.  Proportion  is  an  equality  of  ratios.    Thus,  the  ratios 
6  :  4  and  12:8,  each  being  equal  to  f ,  form  a  proportion. 

368.  Proportion  is  indicated  in  two  ways. 

1st.    By  a  double  colon  placed  between  the  two  ratios ;  thus, 

2  :  5  : :  4  :  10. 

2d.    By  the  sign  of  equality  placed  between  the  two  ratios ; 
thus,  2  :  5  =  4  :  10. 

369.  Since  each  ratio  consists  of  two  terms,  every  pro- 
portion must  consist  of  at  least  four  terms. 

370.  The  Extremes  are  the  first  and  fourth  terms. 

371.  The  Means  are  the  second  and  third  terms. 

372.  Three  numbers  maybe  in  proportion  when  the  first 
is  to  the  second  as  the  second  is  to  the  third.     Thus,  the  num- 
bers 3,  9,  and  27  are  in  proportion  since  3  :  9  : :  9  :  27,  the 
ratio  of  each  couplet  being  3. 

In  such  a  proportion  the  second  term  is  said  to  be  a  mean 
proportional  between  the  other  two. 

373.  In  every  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means.     Thus,  in  the  proportion 

3  :  5  : :  6  :  10  we  have  3  X  10  =  5  X  <>. 


SIMPLE  PROPORTION.  281 

Four  numbers  that  are  proportional  in  the  direct 
order  are  proportional  by  inversion,  and  also  by  alternation,  or 
by  inverting  the  means.  Thus,  the  proportion  2  :  3  : :  6  :  9, 
by  inversion  becomes  3  :  2  : :  9  :  6,  and  by  alternation  2:6:: 
3:9. 

375.  From  the  preceding  principles  and  illustrations,  it 
follows  that,  any  three  terms  of  a  proportion  being  given,  the 
fourth  may  readily  be  found  by  the  following 

RULE.  I.  Divide  the  product  of  the  extremes  by  one  of  the 
means,  and  the  quotient  will  be  the  other  mean.  Or, 

II.  Divide  the  product  of  the  means  by  one  of  the  extremes, 
and  the  quotient  will  be  the  other  extreme. 

EXAMPLES    FOR    PRACTICE, 

Find  the  term  not  given  in  each  of  the  following  proportions* 

1.  48  :  20  : :  (     )  :  50.  Ans.    120. 

2.  42  :  70  : :  3  :  (     ).  Ans.    5. 

3.  (    )  :  30  : :  20  :  100.  Ans.    6. 

4.  1  :  (    )  ::7  :  84.  Ans.    12. 

5.  48yd.  :(    )::  $67.25  :  $201.75.          Ans.    144yd. 

6.  3  Ib.  12  oz.  :  (     )  : :  $3.50  :  $10.50.     Ans.  11  Ib.  4  oz. 

7.  (    )  :  $38.25  : :  8  bu.  2  pk.  :  76  bu.  2  pk.    Ans.  $4.25. 

8.  4£:  381  ::  (    )  :  76£.  Ans.   8±. 

9.  (     )  :  12  : :  |  :  If .  Ans.    7. 
10.  A:(    )::J:f.  Ans.   f. 

SIMPLE  PROPORTION. 

376.  Simple  Proportion  is  an  equality  of  two   simple 
ratios,  and  consists  of  four  terms,  any  three  of  which  being 
given,  the  fourth  may  readily  be  found. 

377.  Every  question  in  simple  proportion  involves  the 
principle  of  cause  and  effect. 

378.  Causes   may  be    regarded  as   action,  of  whatever 
kind,  the  producer,  the  consumer,  men,  animals,  time,  distance, 
weight,  goods  bought  or  sold,  money  at  interest,  &c. 

379.  Effects  may  be  regarded  as  whatever  is   accom- 

x* 


282  PROPORTION. 

plished  by  action  of  any  kind,  the  thing  produced  or  consumed, 
money  paid,  &c. 

38®.  Causes  and  effects  are  of  two  kinds  —  simple  and 
compound. 

381.  A  Simple  Cause,  or  Effect,  contains  but  one  element ; 
as  goods  purchased  or  sold,  and  the  money  paid  or  received 
for  them. 

383.  A  Compound  Cause,  or  Effect,  is  the  product  of  two 
or  more  elements ;  as  men  at  work  taken  in  connection  with 
time,  and  the  result  produced  by  them  taken  in  connection 
with  dimensions,  length  and  breadth,  &c. 

383.  Causes  and  effects  that  admit  of  computation,  that 
is,  involve  the  idea  of  quantity,  may  be  represented  by  num- 
bers, which  will  have  the  same  relation  to  each  other  as  the 
things  they  represent.     And  since  it  is  a  principle  of  philoso- 
phy that  like  causes  produce  like  effects,  and  that  effects  are 
always  in  proportion  to  their  causes,  we  have  the  following 
proportions : 

1st  Cause  :  2d  Cause  :  :  1st  Effect  :  2d  Effect. 
Or,  1st  Effect  :  2d  Effect :  :  1st  Cause  :  2d  Cause; 

in  which  the  two  causes,  or  the  two  effects  forming  one  coup- 
let, must  be  like  numbers,  and  of  the  same  denomination. 

Considering  all  the  terms  of  the  proportion  as  abstract  num- 
bers, we  may  say  that 

1st  Cause  :  1st  Effect :  :  2d  Cause  :  2d  Effect, 
which  will  produce  the  same  numerical  result. 

But  as  ratio  is  the  result  of  comparing  two  numbers  or 
things  of  the  same  kind  (364),  the  first  form  is  regarded  as 
the  most  natural  and  philosophical. 

384.  Simple  causes  and  simple  effects  give  rise  to  simple 
ratios ;  compound  causes  and  compound  effects  to  compound 
ratios. 

38«*.     1.  If  5  tons  of  coal  cost  $30,  what  will  3  tons  cost  ? 

NOTE.  The  required  term  will  be  denoted  by  a  (  ),  and  designated 
"blank." 


SIMPLE  PROPORTION. 


283 


STATEMENT. 

tons.      tons.  $  $ 

5     :     3    :  :    30    :    (     ) 
1st  cause.  2d  cause.    1st  effect.  2d  effect. 

OPERATION. 
5  V  (     )  =  3  X  30 

o  v  2/i6 
o  ^  w_ 


ANALYSIS.    In  this 

example  an  effect  is 
required,  and  5  tons 
must  have  the  same 
ratio  to  3  tons,  as 
$30,  the  cost  of  5 
tons,  to  (blank)  dol- 


tons. 


bar. 

15 

1st  cause. 


STATEMENT. 

bar.  $ 

(     )  :  :    90 
2d  cause.     1st  effect. 
OPERATION. 


$ 

30 
2d  effect. 


00 


$0 


Since  the  product  of  the  extremes  is  equal  to  the  product  of  the 
means  (373),  and  the  product  of  the  means  divided  by  one  of  the 
extremes  will  give  the  other;  (blank)  dollars  will  be  equal  to  the 
product  of  3  X  30  divided  by  5,  which  is  $18,  Ans. 

2.    If  15  barrels  of  flour  cost  $90,  how  many  barrels  can  be 

bought  for  $30  ? 

ANALYSIS.  In  this  ex- 
ample a  cause  is  required, 
and  the  statement  may  be 
read  thus:  If  15  barrels 
cost  $90,  how  many  or 
(blank)  barrels  will  cost 
$30?  The  product  of  the 
extremes,  30  X  15,  di- 
(  )  =  5  bar.,  Ans.  vided  by  the  given  mean, 

90,  will  give  the  required 

term,  5,  as  shown  in  the  operation.    Hence  we  deduce  the  following 

RULE.  I.  Arrange  the  terms  in  the  statement  so  that  the 
causes  shall  compose  one  couplet,  and  the  effects  the  other,  put- 
ting (  )  in  the  place  of  the  required  term. 

II.  If  the  required  term  be  an  extreme,  divide  the  product 
of  the  means  by  the  given  extreme  ;  if  the  required  term  be  a 
mean,  divide  the  product  of  the  extremes  by  the  given  mean. 

NOTES.  1.  If  the  terms  of  any  couplet  be  of  different  denominations, 
they  must  be  reduced  to  the  same  unit  value. 

2.  If  the  odd  term  be  a  compound  number,  it  must  be  reduced  to  its 
lowest  unit. 

3.  If  the  divisor  and  dividend  contain  one  or  more  factors  common 
to  both,  they  should  be  canceled.     If  any  of  the  terms  of  a  proportion 
contain  mixed  numbers,  they  should  first  be  changed  to  improper  frac- 
tions, or  the  fractional  part  to  a  decimal. 

4.  "When  the  vertical  line  is  used,  the  divisor  and  the  required  term 
are  written  on  the  left,  and  the  terms  of  the  dividend  on  the  right. 


284  PROPORTION. 

386.  We  will  now  give  another  method  of  solving  ques- 
tions in  simple  proportion,  without  making  the  statement,  and 
which  may  be  used,  by  those  who  prefer  it,  to  the  one  already 
given.  We  will  term  it  the 

SECOND  METHOD. 

Every  question  which  properly  belongs  to  simple  propor- 
tion must  contain  four  numbers,  at  least  three  of  which  must 
be  given  (3 TO).  Of  the  three  given  numbers,  one  must 
always  be  of  the  same  denomination  as  the  required  number. 
The  remaining  two  will  be  like  numbers,  and  bear  the  same 
relation  to  each  other  that  the  third  does  to  the  required  num- 
ber; in  other  words,  the  ratio  of  the  third  to  the  required 
number  will  be  the  same  as  the  ratio  of  the  other  two  num- 
bers. 

Regarding  the  third  or  odd  term  as  the  antecedent  of  the  sec- 
ond couplet  of  a  proportion,  we  find  the  consequent  or  required 
term  by  multiplying  the  antecedent  by  the  ratio  (366). 

By  comparing  the  two  like  numbers,  in  any  given  question, 
with  the  third,  we  may  readily  determine  whether  the  answer, 
or  required  term,  will  be  greater  or  less  than  the  third  term  ; 
if  greater,  then  the  ratio  will  be  greater  than  1,  and  the  two 
like  numbers  may  be  arranged  in  the  form  of  an  improper 
fraction  as  a  multiplier ;  if  the  answer,  or  required  term,  is  to 
be  less  than  the  third  term,  then  the  ratio  will  be  less  tfran  1, 
and  the  two  like  numbers  may  be  arranged  in  the  form  of  a 
proper  fraction,  as  a  multiplier. 

1.   If  4  cords  of  wood  cost  $12,  what  will  20  cords  cost? 

OPERATION.  ANALYSIS.    It  will 

^^3  x  20  De  readily  seen  in  this 

1£  X  -^,  written =  $60.         example,  that  4  cords 

^  and  20  cords  are  the 

like  terms,  and   that 

$12  is  the  third  term,  and  of  the  same  denomination  as  the  answer 
or  required  term. 

If  4  cords  cost  $12,  will  20  cords  cost  more,  or  less,  than  4  cords? 
evidently  more :  then  the  answer  or  required  term  will  be  greater 


SIMPLE  PRO  PORTION.  285 

than.the  third  term,  and  the  ratio  greater  than  1.  The  ratio  of  4 
cords  to  20'  cords  is  ^,  or  5 ;  hence  the  ratio  of  $12  to  the  answer 
must  be  5,  and  the  answer  will  be  -^  or  5  times  $12,  which  is  $60. 

2.   If  12  yards  of  cloth  cost  $48,  what  will  4  yards  cost  ? 

OPERATION.  ANALYSIS.     In  this   example  we 

48  X  T42  =  $16,  Ans.         see  that  12  yards  and  4  yards  are 
the   like   terms   and  $48  the  third 
term,  and  of  the  same  denomination  as  the  required  answer. 

If  12  yards  cost  $48,  will  4  yards  cost  more  or  less  than  12  yards? 
less :  then  the  ratio  will  be  less  than  1,  and  the  multiplier  a  proper 
fraction.  The  ratio  of  12  yards  to  4  yards  is  T^ ;  hence  the  ratio  of 
$48  to  the  answer  is  T\,  and  the  answer  will  be  T^  times  $48,  which 
is  $16.  Hence  the  following 

RULE.  I.  With  the  two  given  numbers,  which  are  of  the 
same  name  or  kind,  form  a  ratio  greater  or  less  than  1,  accord- 
ing as  the  answer  is  to  be  greater  or  less  than  the  third  given 
number. 

II.  Multiply  the  third  number  by  this  ratio,  and  the  product 
will  be  the  required  number  or  answer. 

NOTE.  1.  Mixed  numbers  should  first  be  reduced  to  improper  frac- 
tions, and  the  ratio  of  the  fractions  found  according  to  (365). 

2.  Reductions  and  cancellation  may  be  applied  as  in  the  first  method. 

The  following  examples  may  be  solved  by  either  of  the 
foregoing  methods. 

EXAMPLES    FOR   PRACTICE. 

1.  If  48  cords  of  wood  cost  $120,  how  much  will  20  cords 
cost?  Ans.    $50. 

2.  If  6  bushels  of  corn  cost  $4.75,  how  much  will  75  bush- 
els cost  ?  Ans.    $59.37£. 

3.  If  8  yards  of  cloth  cost  $3£,  how  many  yards  can  be 
bought  for  $50  ?  Ans.    114f  yds. 

4.  If  12  horses  consume  42  bushels  of  oats  in  3  weeks,  how 
many  bushels  will  20  horses  consume  in  the  same  ^time  ? 

$.  If  7  pounds  of  sugar  cost  75  cents,  how  many  pounds 
can  be  bought  for  $9  ?  Ans.  84  Ibs. 

6.  What  will  11  Ib.  4  oz.  of  tea  cost,  if  3  Ib.  12  oz.  cost 
$3.50?  Ans.  $10.50. 


286  SIMPLE   PROPORTION. 

7.  If  a  staff  3  ft.  8  in.  long  cast  a  shadow   1  ft.  6  in.,  what 
is  the  height  of  a  steeple  that  casts  a  shadow  75  feet  at  the 
same  time  ?  Ans.    183  ft,  4  in. 

8.  At  $2.75  for  14  pounds  of  sugar,  what  will  be  the  cost 
of  100  pounds?  Ans.    $19.64?. 

9.  How  many  bushels  of  wheat  can  be  bought  for  $51.06, 
if  12  bushels  can  be  bought  for  $13.32  ? 

10.  What  will  be  the  cost  of  28J-  gallons  of  molasses,  if  15 
hogsheads  cost  $236.25  ?  Ans.    $7.12J. 

11.  If  7  barrels  of  flour  are  sufficient  for  a  family  6  months, 
how  many  barrels  will  they  require  for  11  months? 

12.  At  the  rate  of  9  yards  for  £512  s.,  how  many  yards  of 
cloth  can  be  bought  for  £44  16s.?  Ans.    72  yds. 

13.  An  insolvent  debtor  fails  for  $7560,  of  which  he  is 
able  to  pay  only  $3100 ;  how  much  will  A  receive,  whose 
claim  is  $756?  Ans.    $310. 

14.  If  2  pounds  of  sugar  cost  25  cents,  and  8  pounds  of 
sugar  are  worth  5  pounds  of  coffee,  what  will  100  pounds  of 
coffee  cost  ?  4ns.    $20.     ' 

15.  If  the  moon  move  13°  10'  35''  in  1  day,  in  what  time 
will  it  perform  one  revolution  ? 

16.  If  8J  bushels  of  corn  cost  $4.20,  what  will  be  the  cost 
of  13£  bushels  at  the  same  rate  ?  Ans.    $6.48. 

17.  If  1  f  yards  of  cotton  cloth  cost  6£  pence,  how  many 
yards  can  be  bought  for  £10  6  s.  8  d.  ?  Ans.  694f  yds. 

18.  If  12J-  cwt.  of  iron  cost  $42£,  how  much  will  48 1  cwt. 
cost?  Ans.    $163.50-1-. 

19.  What  quantity  of  tobacco  can  be  bought  for  $317.23, 
if  8  J  Ibs.  cost  $lf  ?  Ans.    15  cwt.  22.7+  Ibs. 

20.  If  15|  bushels  of  clover  seed  cost  $156|,  how  much 
can  be  bought  for  $95.75  ?  Ans.    9  bu.  2  pk.  2§  qt, 

21.  If  |  of  a  barrel  of  cider  cost  $fy,  how  much  will  $•  of 
a  barrel  cost  ?  .  *  Ans.    $T9T. 

22.  If  a  piece  of  land  of  a  certain  length,  and  4  rods  in 
breadth,  contain  £  of  an  acre,  how  much  would  there  be  if  it 
were  llf  rods  wide  ?  Ans.    2  A.  28  rods. 

23.  If  12£  cwt.  of  iron  cost  $42|,  what  will  48$  cwt.  cost? 


SDIPLE   PROPORTION.  287 

24.  A  grocer  has  a  false  balance,  by  which  1  pound  will 
weigh  but  12  oz.  ;  what  is  the  real  value  of  a  barrel  of  sugar 
that  he  sells  for  $28  ?  Ans.    $21. 

25.  A  butcher  in  selling  meat  sells  14|£  oz.  for  a  pound  ; 
how  much  does  he  cheat  a  customer,  who  buys  of  him  to  the 
amount  of  $;jo  ?  Ans.    $2.46  -f-  . 

26.  If  a  man  clear  $750  by  his  business  in  1  yr.  6  mo.,  how 
much  would  he  gain  in  3  yr.  9  mo.  at  the  same  rate  ? 

27.  If  a  certain  business  yield  $350  net  profits  in  10  mo., 
in  what  time  would  the  same  business  yield  $1050  profits  ? 

28.  B  and  C  have  each  a  farm  ;  B's  farm  is  worth  $25 
an  acre,  and  C's  $30J-  ;  if  in  trading  B  values  his  land  at  $28 
an  acre,  what  value  should  C  put  upon  his  ?     Ans.    $34.16. 

29.  If  I  borrow  $500,  and  keep  it  1  yr.  4  irio.,  for  how  long 
time  should  I  lend  $240  as  an  eouivalent  for  the  favor  ? 

Ans.    2  yr.  9  mo.  10  da. 

COMPOUND   PROPORTION. 

387.  Compound  Proportion  embraces  that  class  of  ques- 
tions in  which  the  causes,  or  the  effects,  or  both,  are  compound. 

The  required  term  may  be  a  cause,  or  a  single  element  of  a 
cause  ;  or  it  may  be  an  effect,  or  a  single  element  of  an  effect* 

1.  If  16  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ? 


i 


STATEMENT. 
1st  causo.  2d  cause.  1st  effect.      2cl  effect 

ll- 

Or,          16  X  50     :     5  X  90    :  :    128     :     (      ) 

OPERATION.  ANALYSTS.    In  this  ex- 

$  X  009  X  ;f$$8  ample  the  required  term 

—   —   72    bu.  ig  the  second  effect;  and 

X$  X  $0  the  question  may  be  read, 

If  16  horses   in  50  days 

consume  128  bushels  of  oats,  5  horses  in  90  days  will  consume 
how  many,  or  (blank)  bushels  ? 

NOTE.     These  questions  are  most  readily  performed  by  cancellation. 


288  PROPORTION. 

2.    If  $480  gain  $84  interest  in  30  months,  what  sum  will 
gain  $21  in  15  months? 

STATEMENT. 

1st  cause.         2d  cause.         1st  effect.     2d  effect. 

84   •    21 

°*       *        "*- 


(30          (15 

OPERATION.  ANALYSIS.    The  re- 

$02  X  &X.  quired  term  in  this  ex- 

—  —  $240,  Ans.         ample  is  an  element  of 
$^£  X  <!$  the  second  cause ;  and 

the   question    may  be 

read,  If  $480  in  30  months  gain  $84,  what  principal  in  15  months 
will  gain  $21  ? 

3.  If  7  men  dig  a  ditch  60  feet  long,  8  feet  wide,  and  6  feet 
deep,  in  12  days,  what  length  of  ditch  can  21  men  dig  in  2f 
days,  if  it  be  3  feet  wide  and  8  feet  deep  ? 

STATEMENT. 

7       121  (60       (    (     ) 

12  :    I    2f  :  :   {   8  :   •<       3 

(6(8 
Or,  7  X  12  :  21  X  f :  :  60  X  8  X  6  :   (     )  X  3  X  8 

OPERATION.  ANALYSIS.    In 

2X  X     8  X  005  X  $  X  #2  this  example  the 

T"  —  80  ft.,  Ans.     required  term  is 
the  length  of  the 


Or, 


3  ditch,  and  is  an  element  of  the 

^  second    effect.      The    question, 

as  stated,  will  read  thus  :   if  7 

005  men,  in  12  days,  dig  a  ditch  GO 

$  feet  long,   8  feet  wide,  and  6 

/£2  feet  deep,  21  men,  in  2|  days, 

will  dig  a  ditch  how  many,  or 

(blank)  feet  long,  3  feet  wide, 


(      )  —  80  ft.,  Ans.          and  8  feet  deep  ? 

Hence  we  have  the  following 

HULK.  I.  Of  tic  f/iven  terms,  select  tloxc  which  constitute 
ihc  anuses,  and  those  which  constitute  the  effects,  ami  arrange 
them  in  couplets,  putting  (  )  in  place  of  the  required  term. 


COMPOUND   PROPORTION.  289 

II.  Then,  if  the  blank  term  (  )  occur  in  either  of  the  ex- 
tremes, make  the  product  of  the  means  a  dividend,  and  the 
product  of  the  extremes  a  divisor  ;  but  if  the  blank  term  occur 
in  either  mean,  make  the  product  of  the  extremes  a  dividend^ 
and  the  product  of  the  means  a  divisor. 

NOTES.  1.  The  causes  must  be  exactly  alike  in  the  number  and  kind 
of  their  terms  ;  the  same  is  true  of  the  effects. 

2.  The  same  preparation  of  the  terms  by  reduction  is  to  be  observed 
as  in  simple  proportion. 

888.  We  will  now  solve  an  example  according  to  the 
Second  Method  given  in  Simple  Proportion. 

1.  If  18  men  can  build  42  rods  of  wall  in  16  days,  how 
many  men  can  build  28  rods  in  8  days  ? 

OPERATION.  ANALYSIS.  We  see  in 

$$4  ^02  tnis  example  that  all  the 

Xffi  X  —  X =  24  men.  terms  appear  in  couplets, 

4$  except  one,  which  is  18 

men,  and  that  is  of  the  same  kind  as  the  required  answer. 

Since  compound  proportion  is  made  up  of  two  or  more  simple 
proportions,  if  this  third  or  odd  term  be  multiplied  by  the  compound 
ratio,  or  by  the  simple  ratio  of  each  couplet  successively,  the  prod- 
uct will  be  the  required  term. 

By  comparing  the  terms  of  each  couplet  with  the  third  term  we 
may  readily  determine  whether  the  answer,  or  term  sought,  will  be 
greater  or  less  than  the  third  term ;  if  greater,  then  the  ratio  will  be 
greater  than  1,  and  the  multiplier  an  improper  fraction  ;  if  less,  the 
ratio  will  be  less  than  1,  and  the  multiplier  a  proper  fraction. 

First  we  will  compare  the  terms  composing  the  first  couplet,  42 
rods  and  28  rods,  with  the  third  term,  18  men.  If  42  rods  require 
18  men,  how  many  men  will  28  rods  require?  less  men ;  hence  the 
ratio  is  less  than  1,  and  the  multiplier  a  proper  fraction,  |f  ;  next, 
if  16  days  require  18  men,  how  many  men  will  8  days  require  ? 
more  men  ;  hence  the  ratio  is  greater  than  1,  and  the  multiplier  an 
improper  fraction,  ^¥6-.  Regarding  the  third  term  as  the  antecedent 
of  a  couplet,  the  consequent  being  the  term  sought,  if  we  multiply 
this  third  term  by  the  simple  ratios,  or  by  their  product,  we  shall 
have  the  required  term  or  answer,  thus :  18  X  ff  X  Y  —  24,  as 
shown  in  the  operation. 

2.  5  compositors,  in  16  days,  of  14  hours  each,  can  compose 
20  sheets  of  24  pages  in  each  sheet,  50  lines  in  a  page,  and 


290  COMPOUND   PROPORTION. 

40  letters  in  a  line ;  in  how  many  days,  of  7  hours  each,  will 
10  compositors  compose  a  volume  to  be  printed  in  the  same 
letter,  containing  40  sheets,  16  pages  in  a  sheet,  60  lines  in  a 
page,  and  50  letters  in  a  line  ?  Ans.  32  days. 

OPERATION, 
days.    comp.    hours,   sheets,    pages,    lines.    lottos. 

16X  A  X  -V4  X  f&  X  |}  X  f  %  X  1%  =  32  days. 

BY  CANCELLATION.          ANALYSIS.     The  required  term  or  an- 
16  —  swer  is  to  be  in  days ;  and  we  see  that 

/£  all  the  terms  appear  in  pairs  or  couplets, 

jj  except  the  16  days,  which  is  of  the  same 

kind  as  the  answer  sought. 

40  We  will  proceed  to  compare  the  terms 

of  each  couplet  with  the  16  days.  First, 
if  5  compositors  require  16  days,  how 
many  days  will  10  compositors  require  ? 
less  days;  hence  the  multiplier  is  the 
32  days,  Ans.  proper  fraction  T5^,  and  we  have  16  X  yV 
Next,  if  14  hours  a  day  require  16  days, 

how  many  days  will  7  hours  a  day  require  ?  more  days ;  hence  the 
multiplier  is  the  improper  fraction  ^,  and  we  have  16  X  ^  X  V' 
Next,  if  20  sheets  require  16  days,  how  many  days  will  40  sheets 
require  ?  move  days ;  hence  the  multiplier  is  the  improper  fraction 
£$,  and  we  have  16  X  Y%  X  ty  X  ff •  Pursuing  the  same  method 
with  the  other  couplets,  we  obtain  the  result  as  shown  in  the  opera- 
tion. Hence  we  have  the  following 

RULE.  I.  Of  the  terms  composing  each  couplet  form  a 
ratio  greater  or  less  than  1,  in  the  same,  manner  as  if  the 
answer  depended  on  those  two  and  the  third  or  odd  term. 

II.  Multiply  the  third  or  odd  term  by  these  ratios  successively, 
and  the  product  will  be  the  answer  sought. 

NOTE.  By  the  odd  term  is  meant  the  one  that  is  of  the  same  kind 
as  the  answer. 

The  following  examples  may  be  solved  by  either  of  the 
given  methods. 

EXAMPLES    FOR    PRACTICE. 

1.  If  16  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ? 


COMPOUND   PROPORTION.  291 

2.  If  a  man  travel  120  miles  in  3  days  when  the  days  are 
12  hours  long,  in   how  many  days  of  10  hours  each  will  he 
require  to  travel  360  miles  ?  Ans.  lOf  days. 

3.  If  6  laborers  dig  a  ditch  34  yards  long  in  10  days,  how 
many  yards  can  20  laborers  dig  in  15  days?     Ans.  170  yds. 

4.  If  450  tiles,  each  12  inches  square,  will  pave  a  cellar, 
how  many  tiles  that  are  9  inches  long  and  8  inches  wide  will 
pave  the  same  ?  Ans.    900. 

5.  If  it  require  1200  yards  of  cloth  f-  wide  to  clothe  500 
men,  how  many  yards  which  is  £  wide  will  it  take  to  clothe 
960  men  ?  Ans.    3291f  yds. 

6.  If  8  men  will  mow  36  acres  of  grass  in  9  days,  of  9 
hours  each  day,  how  many  men  will  be  required  to  mow  48 
acres  in  12  days,  working  12  hours  each  day  ?       Ans.  6  men. 

7.  If  4  men,  in  2£  days,  mow  6|  acres  of  grass  by  work- 
ing 8£  hours  a  day,  how  many  acres  will  15  men  mow  in  3j 
days  by  working  9  hours  a  day?  Ans.   40}^  acres. 

8.  If,  by  traveling  6  hours  a  day  at  the  rate  of  4£  miles 
an  hour,  a  man  perform  a  journey  of  540  miles  in  20  days,  in 
how  many  clays,  traveling  9  hours  a  day  at  the  rate  of  4§ 
miles  an  hour,  will  he  travel  600  miles  ?       Ans.    14f  days. 

9.  If  21  yards  of  cloth  If  yards  wide  cost  $3.37£,  what 
cost  361  yards,  11  yards  wide?  Ans.    $52.79  +. 

10.  If  5  men  reap  52.2  acres  in  6  days,  how  many  men 
will  reap  417.6  acres  in  12  days  ?  Ans.    20  men. 

11.  If  6  men  dig  a  cellar  22.5  feet  long,  17.3  feet  wide,  and 
10.25  feet  deep,  in  2.5  days,  of  12.3  hours,  in  how  many  days, 
of  8.2  hours,  will  9  men  take  to  dig  another,  measuring  45 
feet  long,  34.6  wide,  and  12.3  deep?  Ans.    12  days. 

12.  If  54  men  can  build  a  fort  in  244  days,  working  12£ 
hours  each  day,  in  how  many  days  will  75  men  do  the  same, 
when  they  work  but  101  hours  each  day  ?        Ans.  21  days. 

13.  If  24  men  dig  a  trench  33f  yards  long,  5-|  wide,  and 
3J-  deep,  in  189  days,  working  14  hours  each  day,  how  many 
hours  per  day  must  217  men  work,  to  dig  a  trench  23 1  yards 
long,  3§  wide,  and  2£  deep,  in  5^  days  ?       Ans.    16  hours. 


292  PARTNERSHIP. 


PARTNERSHIP. 

389.  Partnership  is  a  relation  established  between  two 
or  more  persons  in  trade,  by  which  they  agree  to  share  the 
profits  and  losses  of  business. 

390.  The  Partners  are  the  individuals  thus  associated. 

391.  Capital,  or  Stock,  is  the  money  or  property  invested 
in  trade. 

392.  A  Dividend  is  the  profit  to  be  divided. 

393.  An  Assessment  is  a  tax  to  meet  losses  sustained. 

CASE   I. 

394.  To  find  each  partner's  share  of  the  profit  or 
loss,  when  their  capital  is  employed  for  equal  periods  of 
time. 

1.  A  and  B  engage  in  trade;  A  furnishes  $300,  and  B 
$400  of  the  capital  $  they  gain  $182;  what  is  each  one's 
share  of  the  profit  ? 

OPERATION.  ANALYSIS.  Since 

$300  the    whole    capital 

$4.00  employed  is  $300 

+  $400  =  $700,  it 

$700,  whole  stock.  is   evident  that  A 

f  -g.o  —  .3,  A's  share  of  the  stock.  furnishes  f  £-§-  =z  f 

4°-°  —  4    B»g     «       «        «  of  the  capital,  and 

$182  X  ?'  =  $78,  A's  share  of  the  gain.  B  *#f  =/  °f  .the 

capital.  And  since 

$182X*  =  *104,B-.  -  «  «  each  man's  share  of 

the  profit  or  loss  will  have  the  same  ratio  to  the  whole  profit  or 
loss  that  his  share  of  the  stock  has  to  the  whole  stock,  A  will  have 
-^  of  the  entire  profit,  and  B  ^,  as  shown  in  the  operation. 

We  may  also  regard  the  whole  capital  as  the  first  cause, 
and  each  man's  share  of  the  capital  as  the  second  cause,  the 
whole  profit  or  loss  as  the^rs^  effect,  and  each  man's  share  of  the 
profit  or  loss  as  the  second  effect,  and  solve  by  proportion  thus : 


PARTNERSHIP. 


1st  cause.        2<1  cause. 

$700    :    $300 
$700    : 


#00 
(     ) 


B,s  profit. 


(      )  —  <J>  I  O,    A'S  profit. 

Hence  we  have  the  following 

RULE.  Multiply  the  whole  profit  or  loss  by  the  ratio  of  each 
man's  share  of  the  capital  to  the  whole  capital.  Or, 

The  whole  capital  is  to  each  man's  share  of  the  capital  as  the 
whole  profit  or  loss  is  to  each  man's  share  of  the  profit  or  loss. 

.2  Three  men  trade  in  company;  A  furnishes  $8000,  B 
$12000,  and  C  20000  of  the  capital;  their  gain  is  $1680; 
what  is  each  man's  share  ? 

Ans.  A's  $336;  B's  $504;  C's  $840. 

3.  Three  persons  purchased  a  house  for  $2800,  of  which  A 
paid  $1200,  B  $1000,  and  C  $600  ;   they  rented  it  for  $224 
a  year  ;  how  much  of  the  rent  should  each  receive  ? 

4.  A  man  failed  in  business  for  $20000,  and  his  available 
means  amounted  to  only  $13654  ;  how  much  will  two  of  his 
creditors  respectively  receive,  to  one  of  whom  he  owes  $3060, 
and  to  the  other  $1530  ?         Ans.  $2089.062  ;  $1044.531. 

5.  Four  men  hired  a  coach  for  $13,  to  convey  them  to 
their  respective  homes,  which  were  at  distances  from  the  place 
of  starting  as  follows  :    A's  16  miles,  B's  24  miles,  C's  28 
miles,  and  D's  36  miles;   what  ought  each  to  pay? 


. 
S' 


(  A  $2.      C.  $3.50. 
'(B$3.     D$4.50. 

6.  A  captain,  mate,  and  12  sailors  took  a  prize  of  $2240, 
of  which  the  captain  took  14  shares,  the  mate  6  shares,  and 
each  sailor  1  share  ;  how  much  did  each  receive  ? 

7.  A  cargo  of  corn,  valued  at  $3475.60,  was  entirely  lost  ; 
£  of  it  belonged  to  A,  £  of  it  to  B,  and  the  remainder  to  C  ; 
how  much  was  the  loss  of  each,  there  being  an  insurance  of 
$2512?     Ans.  $120.45,  A's.     $240.90,  B's.     $602.35,  C's. 

Y* 


294  PARTNERSHIP. 

8.  Three  persons  engaged  in  the  lumber  trade ;  two  of  the 
persons  furnished  the  capital,  and  the  third  managed  the  busi- 
ness; they  gained  $2571.24,  of  which  C  received  $6  as  often 
as  D  $4,  and  E  had  -^  as  much  as  the  other  two  for  taking  care 
of  the  business  ;  how  much  was  each  one's  share  of  the  gain  ? 

.Am.  $1285.62,  C's.     $857.08,  D's.     $428.54,  E's. 

9.  Four  persons    engage   in  the   coal   trade;    D  puts  in 
$3042  capital ;  they  gain  $7500,  of  which  A  takes  $2000,  B 
$2800.75,  and  C  $1685.25  ;  how  much  capital  did  A,  B,  and 
C  put  in,  and  how  much  is  D's  share  of  the  gain  ? 

A       (A,  $6000.          C,  $5055.75. 
s'  I  B,  $8402.25.     D's  gain,  $1014. 

CASE   II. 

395.  To  find  each  partner's  share  of  the  profit  or 
loss  when  their  capital  is  employed  for  unequal  periods 
of  time. 

It  is  evident  that  the  respective  shares  of  profit  and  loss  will 
depend  upon  two  conditions,  viz. :  the  amount  of  capital  in- 
vested by  each,  and  the  time  it  is  employed. 

1.  Two  persons  form  a  partnership;  A  puts  in  $450  for 
7  months,  and  B  $300  for  9  months ;  they  lose  $156 ;  how 
much  is  each  man's  share  of  the  loss  ? 

OPERATION.  ANALYSIS.     The 

$450  X  7  =  $3150,  A's  capital  for  1  mo.  use  of  $450  capital 

$300  X  9  =  $2700,  B'8     «      «     «  for  7  months  is  the 

~ same  as  the  use  of 

$5800,  entire  «        «      «  7    timeg     $450>    or 

f  J$&  =  ^  A's  share  of  the  entire  capital.  $3150  for  l  month5 

nn  =  A    B's      «       «         «  «  and  °f   $300  for  9 

ii  w  v  7      ouu  ,  months  is  the  same 

$156  X  -ft  =  $84,  A'S  loss.  as  the  uge  of  9  timeg 

$156  X  A  =  $72,  B's    «  $300,  or  $2700  for 

1  month.  The  en- 
tire capital  for  1  month  is  equivalent  to  $3150  +  $2700=  $r>«sr>(). 
If  the  loss,  $156,  be  divided  between  the  two  partners,  according  to 
Case  I,  the  results  will  be  the  loss  of  each  as  shown  in  the  operation. 


PARTNERSHIP.  295 

Examples   of  this   kind   may  also  be  solved  by  proportion  as  in 
Case  I,  the  causes  being  compounded  of  capital  and  time ;  thus, 

$5850  :  $3150  :  :  $156  :  (  ) 
$5850  :  $2700  :  :  $156  :  (  ) 


(      ) 


(       )  =  $84,  A's  loss.  (       )  =  $72,  B's  loss. 

Hence  the  following 

RULE.  Multiply  each  man's  capital  by  the  time  it  is  em- 
ployed in  trade,  and  add  the  products.  Then  multiply  the 
entire  profit  or  loss  by  the  ratio  of  each  product  to  the  sum  of 
the  products,  and  the  results  will  be  the  respective  shares  of 
profit  or  loss  of  each  partner.  Or, 

Multiply  each  man's  capital  by  the  time  it  is  employed  in 
trade,  and  regard  each  product  as  his  capital,  and  the  sum  of 
the  products  as  the  entire  capital,  and  solve  by  proportion,  as  in 
Case  I. 

EXAMPLES    FOR    PRACTICE. 

2.  Three  persons  traded  together;   B  put  in  $250  for  6 
months,  C  $275  for  8  months,  and  D  $450  for  4  months; 
they  gained  $795 ;  how  much  was  each  man's  share  of  the 
gain? 

3.  Two  merchants  form  a  partnership  for  18  months.     A  at 
first  put  in  $1000,  and  at  the  end  of  8  months  he  put  in  $600 
more ;  B  at  first  put  in  $1500,  but  at  the  end  of  4  months  he 
drew  out  $300 ;  at  the  expiration  of  the  time  they  found  that 
they  had  gained  $1394.64;  how  much  was  each  man's  share 
of  the  gain?  Ans.   A's  $715.20  ;  B's  $679.44. 

4.  Three  men  took  a  field  of  grain  to  harvest  and  thresh 
for  I  of  the  crop ;    A  furnished  4  hands  5  days,  B  3  hands 
6  days,  and  C  6  hands  4  days  ;   the  whole  crop  amounted  to 
372  bushels ;  how  much  was  each  one's  share  ? 

5.  William   Gallup  began  trade  January  1,  1856,  with  a 
capital  of  $3000,  and,  succeeding  in  business,  took  in  M.  H. 
Decker  as  a  partner  on  the  first  day  of  March  following,  with 


296  ANALYSIS. 

a  capital  of  $2000 ;  four  months  after  they  admit  J.  New- 
man as  third  partner,  who  put  in  $1800  capital;  they  con- 
tinue their  partnership  until  April  1,  1858,  when  they  find 
that  $4388.80  has  been  gained  since  Jan.  1,  1856;  how 
much  was  each  one's  share  ?  (  $2106,  Gallup's. 

Ans.\  $1300,  Decker's. 

(  $  982.80,  Newman's. 

6.  Two  persons  engage  in   partnership  with   a  capital  of 
$5600;   A's  capital  was  in  trade  8  months,  and  his  share  of 
the  profits  was   $560  ;    B's  capital  was  in  10  months,  and  his 
share  of  the  profits  was  $800 ;   what  amount  of  capital  had 
each  in  the  firm  ?  Ans.  A,  $2613.33£;  B,  $2986.66$. 

7.  A,  B,  and  C,  engage  in  trade  with  $1930  capital;   A's 
money  was  in  3  months,  B's  5,  and  C's  7  ;  they  gained   $117, 
which  was  so  divided  that  J-  of  A's  share,  was  equal  to  £  of 
B's  and  to  £  of  C's ;   how  much  did  each  put  in,  and  each 
gain  ?  r  A,  $700  capital ;  $26  gain. 

Ans.  1  B,  $630       «       $39     " 

( C,  $600       «       $52     « 

* 

ANALYSIS.    ' 

396.  Analysis,  in  arithmetic,  is  the  process  of  solving 
problems  independently  of  set  rules,  by  tracing  the  relations 
of  the  given  numbers  and  the  reasons  of  the  separate  steps  of 
the  operation  according  to  the  special  conditions  of  each  question. 

397.  In  solving  questions  by  analysis,  we  generally  reason 
from  the  given  number  to  unity,  or  1,  and  then  from  unity,  or 
1,  to  the  required  number. 

398.  United  States  money  is  reckoned  in  dollars,  dimes, 
cents,andmills(18O),one  dollar  being  uniformly  valued  in  all 
the  States  at  100  cents  ;  but  in  most  of  the  States  money  is 
sometimes  still  reckoned  in  pounds,  shillings,  and  pence. 

NOTE.  At  the  time  of  the  adoption  of  our  decimal  currrnry  by 
Congress,  in  1786,  the  colonial  currency,  or  bills  of  credit,  issued  by  the 
colonies,  had  depreciated  in  value,  and  this  depreciation,  being  unequal 
in  the  different  colonies,  gave  rise  to  the  different  values  of  the  State 
currencies  ;  and  this  variation  continues  wherever  the  denominations  of 
shillings  and  pence  are  in  use. 


ANALYSIS.  297 

399.    In  New  England,  Indiana,  \ 
Illinois,  Missouri,  Virginia,  Kentucky,  >  $1  =  6  s.  —  72  d. 
Tennessee,  Mississippi,  Texas, ) 

New  York,  Ohio,  Michigan, $1  =  8  s.  =  96  d. 

New  Jersey,  Pennsylvania,  Dela-) 
ware,  Maryland, }  f  1  =  7s.  6d.  =  90  d. 

South  Carolina,   Georgia,  Canada,  ) 
Nova  Scotia, J  $1  =  5  *•  =  60  d' 


EXAMPLES    FOR    PRACTICE. 

1.   What  will  be  the  cost  of  42  bushels  of  oats,  at  3  shillings 
per  bushel,  New  England  currency  ? 

OPERATION.  ANALYSIS.    Since 

$  1  bushel  costs  3  shil- 


42  X  3=  126s. 
126-:- 6  =$21     Or, 


lings,  42  bushels  will 
cost  42  times  3s.,  or 


WI,  Ans.         42X3=  126s.;  and 
as  6  s.  make  1  dollar 

New  England  currency,  there  are  as  many  dollars  in  126  s.  as  6  is 
contained  times  in  126,  or  $21. 

2.   What  will   180  bushels  of  wheat  cost  at  9s.  4d.  per 
bushel,  Pennsylvania  currency  ? 

OPERATION.  ANALYSIS.      Multi- 

Or, 


plying  the  number  of 
28  bushels  by  the  price, 

and  dividing  by  the 
jf  _  value  of  1  dollar  re- 


Ans  duced  to  pence,  we 
we  have  $224.  Or, 
when  the  pence  in  the 

given  price  is  an  aliquot  part  of  a  shilling,  the  price  may  be  reduced 
to  an  improper  fraction  for  a  multiplier,  thus  :  9  s.  4  d.  =  9^  s.  = 
^  s.,  the  multiplier.  The  value  of  the  dollar  being  7  s.  6  d.  =  7-J  s. 
=  ^,  we  divide  by  ^-,  as  in  the  operation. 

3.    What  will  be  the  cost  of  3  hhd.  of  molasses,  at  1  s.  3  d. 
per  quart,  Georgia  currency  ? 


298  ANALYSIS. 

OPERATION. 

ANALYSIS.     In  this  example  we  first 
reduce  3  hhd.  to  quarts,  by  multiplying 
by  63  and  4,  and  then  multiply  by  the 
^  price,  either  reduced  to  pence  or  tt>  an 

_*° improper  fraction,   and  divide    by  the 

405  00  value  of  1  dollar  reduced  to  the  same 
denomination  as  the  price. 

$202.50 

4.  Sold  9  firkins  of  butter,  each  containing  56  lb.,  at  1  s.  6  d. 
per  pound,  and  received  in  payment  carpeting  at  6  s.  9  d.  per 
yard  ;  how  many  yards  of  carpeting  would  pay  for  the  butter  ? 

OPERATION.  ANALYSIS.     The  operation  in  this  is  similar 

to  the  preceding  examples,  except  that  we  di- 
vide the  cost  of  the  butter  by  the  price  of  a, 


,  - 


unit  of  the  article  received  in  payment,  reduced 
to  the  same  denominational  unit  as  the  price 


112  yd.          °^  a  un^  °f  ^6  article  sold.    The  result  will  be 
the  same  in  whatever  currency. 

5.  What  will   3    casks   of  rice   cost,  each  weighing    126 
pounds,  at  4  d.  per  pound,  South  Carolina  currency  ?   Ans.  $27. 

6.  How  many  pounds  of  tea,  at  7  s.  per  pound,  must  be 
given  for  28  lb.  of  butter,  at  1  s.  7  d.  per  pound  ?     Ans.    6^. 

7.  Bought  2  casks  of  Catawba  wine,  each  containing  72 
gallons,  for  $648,  and  sold  it  at  the  rate  of  10  s.  6  d.  per  quart, 
Ohio  currency  ;  how  much  was  my  whole  gain  ?     Ans.    $108. 

8.  What  will  be  the  expense  of  keeping  2  horses  3  weeks 
if  the  expense  of  keeping  1  horse  1  day  be  2  s.  6  d.,  Canada 
currency?  Ans.    $21. 

9.  How  many  days'  work,  at  6  s.  3  d.  per  day,  must  be 
given  for  20  bushels  of  apples  at  3  s.  per  bushel  ?     Ans.    9^. 

10.  Bought  160  lb.  of  dried  fruit,  at  1  s.  6d.  a  pound,  in 
New  York,  and  sold  it  for  2  s.  a  pound  in  Philadelphia ;  how 
much  was  my  whole  gain  ?  Ans.   $12.66§. 

11.  A  merchant  exchanged  43£  yards  of  cloth,  worth  10s. 
6  d.  per  yard,  for  other  cloth  worth  8  s.  3  d.  per  yard ;  how 
many  yards  did  he  receive  ?  Ans. 


ANALYSIS.  299 

12.  What  will  be  the  cost  of  300  bushels  of  wheat  at  9  s. 
4  d.  per  bushel,  Michigan  currency  ?  Ans.    $350. 

13.  If  £  of  f  of  a  ton  of  coal  cost  $2f ,  how  much  will  £  of 
6  tons  cost  ? 

OPERATION. 

ANALYSIS..    Since  f  of  4  =:£|  of  a  ton  costs 
$\2-,  1  ton  will  cost  28  times  -^  of  $J£, 
<*  $J£  X  -ff  5  and  f  of  6  tons  nr  Sf.  tons,  will 
cost  -3°  times      .  o 


$16, 

14.  If  8  men  can   build  a  wall  20  ft.  long,  6  ft.  high,  and 
4  ft.  thick,  in  12  days,  working   10  hours  a  day,  in  how  many 
days  can  24  men  build  a  wall  200  ft.  long,  8  ft.  high,  and  6  ft. 
thick,  working  8  hours  a  day  ? 

OPERATION. 

It        1         10       20010       $        0 
-  X—  X- X-     -X  —  X— =100  da. 
I       t&       9         20        0      4 

ANALYSIS.  Since  8  men  require  12  days  of  10  hours  each  to 
build  the  wall,  1  man  would  require  8  times  12  days  of  10  hours 
each,  and  10  times  (12  X  8)  days  of  1  hour  each.  To  build  a  wall 
1  ft.  long  would  require  -fa  as  much  time  as  to  build  a  wall  20  ft. 
long  ;  to  build  a  wall  1  ft.  high  would  require  \  as  much  time  as  to 
build  a  wall  6  ft.  high ;  to  build  a  wall  1  ft.  thick,  J  as  much  time  as 
to  build  a  wall  4  ft.  thick.  Now,  24  men  could  build  this  wall  in  ^ 
as  many  days,  by  working  1  hour  a  day,  as  1  man  could  build  it, 
and  in  |  as  many  days  by  working  8  hours  a  day,  as  by  working  1 
hour  a  day  ;  but  to  build  a  wall  200  ft.  long  would  require  200  times 
as  many  days  as  to  build  a  wall  1  ft.  long ;  to  build  a  wall  8  ft.  high 
would  require  8  times  as  many  days  as  to  build  a  wall  1  ft.  high ; 
and  to  build  a  wall  6  ft.  thick  would  require  6  times  as  many  days 
as  to  build  a  wall  1  ft.  thick. 

15.  If  2  pounds  of  tea  are  worth  11  pounds  of  coffee,  and 
3  pounds  of  coffee  are  worth  5  pounds  of  sugar,  and  18  pounds 
of  sugar  are  worth  21  pounds  of  rice,  how  many  pounds  of 
rice  can  be  purchased  with  12  pounds  of  tea? 


300  ANALYSIS. 


OPERATION.  ANALYSIS.     Since  18  lb.  of  su- 

gar  are  equal  in  value  to  21  Ib.  of 
rice,  1  Ib.  of  sugar  is  equal  to  ^ 
of  21  Ib.  of  rice,  or  fl  —  |  Ib.  of 
rice,  and  5  Ib.  of  sugar  are  equal 
to  5  times  ^  Ib.  of  rice,  or  ^  Ib. ; 


3  |  385  if  3  Ib.  of  coffee  are  equal  to  5  Ib. 

Am     128Mb  °f  sugar'  °r  V  lb-  of  rice>  !  lb-  of 

coffee  is  equal  to  J   of  ^  lb.  of 

rice,  or  f|  lb.,  and  11  lb.  of  coffee  are  equal  to  11  times  f  |  lb.  of 
rice,  or  -3j8¥5-  lb. ;  if  2  lb.  of  tea  are  equal  to  11  lb.  of  coffee,  or  -^g5  lb. 
of  rice,  1  lb  of  tea  is  equal  to  J  of  »f£  lb.  of  rice,  or  -^  }D.,  and  12 
lb.  of  tea  are  equal  to  12  times  ^  lb.  of  rice,  or  a|-&  lb.  =  1281  lb. 

16.  If  16  horses  consume  128   bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days?  Ans.  72. 

17.  If  $10^   wiH  buy  4f  cords   of  wood,  how  many  cords 
can  be  bought  for  $24£  ?  Ans.  11. 

18.  Gave  52  barrels  of  potatoes,  each  containing  3  bushels, 
worth  33^  cents  a  bushel,  for  65  yards  of  cloth ;  how  much 
was  the  cloth  worth  per  yard  ?  Ans.  $.80. 

19.  If  a  staff  3  ft.  long  cast  a  shadow  5  ft.  in  length,  what 
is  the  height  of  an  object  that  casts  a  shadow  of  46f  ft.  at  the 
same  time  of  day?  Ans.  28  ft. 

20.  Three  men  hired  a  pasture  for  $63  ;  A  put  in  8  sheep 
7£  months,  B  put  in   12   sheep  4£  months,  and  C  put  in  15 
sheep  6|  months  ;  how  much  must  each  pay  ? 

21.  If  7  bushels   of  wheat  are  worth  10  bushels  of  rye, 
and  5   bushels  of  rye   are  worth  14  bushels  of  oats,  and  6 
bushels  of  oats  are  worth  $3,  how  many  bushels  of  wheat  will 
$30  buy?  Ans.  15. 

22.  If  $480  gain  $84  in  30  months,  what  capital  will  gain 
$21  m  15  months  ?  Ans.  $240. 

23.  How  many  yards  of  carpeting  f  of  a  yard  wide  are 
equal  to  28  yards  J  of  a  yard  wide?  Ans.  31£. 

24.  If  a  footman  travel  130  miles  in  3  days,  when  the  days 
are  14  hours  long,  in  how  many  days  of  7  hours  each  will  he 
travel  390  miles  ?  Ans.  IS. 


ANALYSIS.  301 

25.  If  6  men  can  cut  45  cords  of  wood  in  3  days,  how 
many  cords  can  8  men  cut  in  9  days  ?  Ans.  180. 

26.  B's  age  is  1^-  times  the  age  of  A,  and  C's  is  2^  times 
the  age  of  both,  and  the  sum  of  their  ages  is  93 ;  what  is  the 
age  of  each?  Ans.     A's  age,  12  yrs. 

27.  If  A  can  do  as  much  work  in  3  days  as  B  can  do  in 
4£  days,  and  B  can  do  as  much  in  9   days  as  C  in  12  days, 
and  C  as  much  in   10  days  as  D  in  8,  how  many  days'  work 
done  by  D  are  equal  to  5  days'  done  by  A?  Ans.   8. 

28.  The  hour  and  minute  hands  of  a  watch  are  together  at 
12  o'clock,  M. ;  when  will  they  be  exactly  together  the  third 
time  after  this  ? 

OPERATION.  ANALYSIS.     Since 

12  X  TIT  X  3  —  3T3T  h.  the  minute  hand  pass- 

Ans.   3h.  16 miri.  21T9Tsec.,  P.M.  es  the  hour  hand  11 

times  in   12  hours,  if 

both  are  together  at  12,  the  minute  hand  will  pass  the  hour  hand 
the  first  time  in  ^  of  12  hours,  or  l^j  hours  ;  it  will  pass  the  hour 
hand  the  second  time  in  T2T  of  12  hours,  and  the  third  time  in  ^  of 
12  hours,  or  3^  hours,  which  would  occur  at  16  min.  21T^  sec. 
past  3  o'clock,  P.  M. 

29.  A  flour  merchant  paid  $164  for  20  barrels  of  flour, 
giving  $9  for  first  quality,  and  $7  for  second  quality ;  how 
many  barrels  were  there  of  each  ? 

OPERATION.  ANALYSIS.     If  all  had  been 

$9  X  20  —  $180 ;  first  quality,  he  would  have  paid 

$180  —  $164  =  $16.  $180»  or  $16  more  than  he  did 

<£C) $ 7  —  $9  .  Pay-     Every  barrel  of  second 

quality  made  a  difference  of  $2 
16-1-2=8  bbl.,  *  «**,.        I  th(f  cost .  hence  there  were 

20  —  8  —  12  bbl.,  1st     «  as  many  barrels  of  second  qual- 

ity as  $2,  the  difference  in  the 
cost  of  one  barrel,  is  contained  times  in  $16,  &c. 

30.  A  boy  bought  a  certain  number  of  oranges  at  the  rate 
of  3  for  4  cents,  and  as  many  more  at  the  rate  of  5  for  8  cents ; 
he  sold  them  again  at  the  rate  of  3  for  8  cents,  and  gained  on 
the  whole  75  cents ;  how  many  oranges  did  he  buy  ? 

z 


302  ANALYSIS. 

OPERATION.  ANALYSIS.   For 

I  +  t  =tt;ft-^2  =  i!>averasecost.       those   he    bought 
£  —  2|  —  i|  —  i£  ctgej  gain  on  each.  at  the  rate  of  3  for 

75  -f-  U  =  90,  number  of  oranges.  4  Cents  he  Paid  $ 

of  a  cent  each,  and 

for  those  he  bought  at  the  rate  of  5  for  8  cents  he  paid  f  of  a  cent 
each ;  and  |  -f  |  —  ||  cents,  what  he  paid  for  1  of  each  kind, 
which  divided  by  2  gives  f  f  cents,  the  average  price  of  all  he  bought. 
He  sold  them  at  the  rate  of  3  for  8  cents,  or  |  cents  each  ;  the  dif- 
ference between  the  average  cost  and  the  price  he  sold  them  for,  or 
f  —  f  f  —  if  =  H  cents,  is  the  gain  on  each  ;  and  he  bought  as 
many  oranges  as  the  gain  on  one  orange  is  contained  times  in  the 
whole  gain,  £c. 

31.  A  man  bought  10  bushels  of  wheat  and  25  bushels  of 
corn  for  $30,  and  12  bushels  of  wheat  and  5  bushels  of  corn 
for  $20 ;  how  much  a  bushel  did  he  give  for  each  ? 

OPERATION.  ANALYSIS.     We  may  divide  or 

w«      C.  multiply  either  of  the  expressions 

1st  lot,       10     25  $30  by  such  a  number  as  will  render 

2d   "         12       5  $20  one  of  the  commodities  purchased, 

alike  in  both  expressions.    In  this 
1st  -L-  5:^:2        5        So  i  v  •  i      i      c,       , 

example  we  divide  the  first  by  5 

10 $14  to  make  the  numbers   denoting 

lbu,w.  :=  $1.40  the  corn  alike,  (the  same  result 

1  bu.  C.  =  $  .64  would  be  produced  by  multiply- 

ing the  second  by  5,)  and  we  have 

the  cost  of  2  bushels  of  wheat  and  5  bushels  of  corn,  equal  to  $6. 
Subtracting  this  from  12  bushels  of  wheat  and  5  bushels  of  corn,  which 
cost  $20,  we  find  the  cost  of  10  bushels  of  wheat  to  be  $14  ;  there- 
fore the  cost  of  1  bushel  is  ^  of  $14,  or  $1.40.  From  any  one  of 
the  expressions  containing  both  wheat  and  corn,  we  readily  find  the 
cost  of  1  bushel  of  corn  to  be  64  cents. 

32.  A,  B,  and   C  agree  to  build  a  barn  for  $270.     A  and 
B  can  do  the  work  in  16  days,  B  and  C  in  13£  days,  and  A 
and  C  in  11^  days.     In  how  many  days  can  all  do  it  working 
together  ?     In  how  many  days  can  each  do  it  alone  ?     What 
part  of  the  pay  ought  each  to  receive  ? 


ANALYSIS. 


303 


OPERATION. 
what  A  and  B  do  in  1  day. 

"   BandC  "     " 

"  Aandc  "     " 

=  **'  what  A'B'and  c  do  in 

2  days. 

1  -f-  2  —  -^-,  what  A,  B,  and  C  do  in  1  day. 
-f-  -g^-  =i  £|  days,  time  A,  B,  and  C,  will  do  the 
whole  work  together. 

-^  =  ^5  1—  ^  =  20  da.,  C  alone. 
9-=3;  l-=:26    da,A  « 


ANALYSIS.  Sinco 
A  and  B  can  do  the 
work  in  16  days,  they 
can  do  fa  of  it  in  1 
day;  B  and  C,  in 
13£  or  ^-  days,  they 
can  do  ^  of  it  in  1 
day;  A  and  C,  in  llf 
or  Sf.  days,  they  can 
do  fa  of  it  in  1  day. 
Then  A,  B,  and  C, 
by  working  2  days 
each,  can  do  fa  -j- 


X  8$-  =  -f ,  the  part  of  the  whole  C  did. 
X  8f  =  f,        "  "          "     A  " 

V  84  =  £ .        «  "         "     B  « 


$270  X  |  =  $120,  C's  share. 

$270Xf  =  $90,    A'S    « 
$270  X  |  =  $60,    B's     « 


work,  and  by  work- 
ing 1  day  each  they 
can  do  -|-  of  -|-|,  or  -^ 
of  the  work  ;  and  it 
will  take  them  as 
many  days  working 
together  to  do  the 
whole  work  as  ¥97  is  contained  times  in  1,  or  8-f-  days. 

Now,  if  we  take  what  any  two  of  them  do  in  1  day  from  what  the 
three  do  in  1  day,  the  remainder  will  be  what  the  third  does ;  we 
thus  find  that  A  does  -g^,  B  -£$,  and  C  -g^-. 

Next,  if  we  denote  the  whole  wrork  by  1,  and  divide  it  by  the  part 
each  does  in  1  day,  we  have  the  number  of  days  that  it  will  take 
each  to  do  it  alone,  viz. :  A  26f  days,  B  40  days,  and  C  20  days. 
And  each  should  receive  such  a  part  of  $270  as  would  be  ex- 
pressed by  the  part  he  does  in  1  day,  multiplied  by  the  number  of 
days  he  works,  which  will  give  to  A  $90,  B  $60,  and  C  $120. 

33.  If  6  oranges  and  7  lemons  cost  33  cents,  and  12  oranges 
and  10  lemons  cost  54  cents,  what  is  the  price  of  1  of  each  ? 

Ans.    Oranges,  2  cents;  lemons,  3  cents. 

34.  If  an  army  of  1000  men  have  provisions  for  20  days, 
at  the  rate  of  18  oz.  a  day  to  each  man,  and  they  be  reinforced 
by  600  men,  upon  what  allowance  per  day  must  each  man  be 
put,  that  the  same  provisions  may  last  30  days  ?    Ans.    7±  oz. 

35.  There  are  54  bushels  of  grain  in  2  bins  ;  and  in  one  bin 
are  6  bushels  less  than  £  as  much  as  there  is  in  the  other ; 
how  many  bushels  in  the  larger  bin  ?  Ans.   40. 


304  ANALYSIS. 

36.  The  sum  of  two  numbers  is  20,  and  their  difference  is 
equal   to   %   of  the   greater    number;    what   is   the   greater 
number?  Am.    12. 

37.  If  A  can  do  as  much  work  in  2  days  as   C  in  3  days, 
and  B  as  much  in  5  days  as  C  in  4  days ;  what  time  will  B 
require  to  execute  a  piece  of  work  which  A  can  do  in  6 
weeks?  Am.    11^  weeks. 

38.  How  many  yards  of  cloth,  f  of  a  yard  wide,  will  line 
36  yards  1£  yards  wide  ?  Am.    60. 

39.  How  many  sacks  of  coffee,  each  containing  104  Ibs , 
at  10  d.  per  pound  N.  Y.  currency,  will  pay  for  80  yards  of 
broadcloth  at  $3£  per  yard?  Am.    24. 

40.  A  person,  being  asked  the  time  of  day,  replied,  the  time 
past  noon  is  equal  to  £  of  the  time  to  midnight ;  what  was 
the  hour  ?  Am.   2,  P.  M. 

41.  A  market  woman  bought  a  number  of  peaches  at  the 
rate  of  2  for  1  cent,  and  as  many  more  at  the  rate  of  3  for  1 
cent,  and  sold  them  at  the  rate  of  5  for  3  cents,  gaining  55 
cents ;  how  many  peaches  did  she  buy  ?  Ans.   300. 

42.  A  can  build  a  boat  in  18  days,  working  10  hours  a  day, 
and  B  can  build  it  in  9  days,  working  8  hours  a  day ;  in  how 
many  days  can  both  together  build  it,  working  6  hours  a  day  ? 

43.  A  man,  after  spending  %  of  his  money,  and  £  of  the 
remainder,  had  $10  left ;    how  much  had  he  at  first  ? 

44.  If  30  men  can  perform  a  piece  of  work  in  1 1  days,  how 
many  men  can  accomplish  another  piece  of  work,  4  times  as 
large,  in  £  of  the  time  ?  Ans.    600. 

45.  If  16£  Ib.  of  coffee  cost  $3£,  how  much  can  be  bought 
for  $1.25?  Ans.    6£lb. 

46.  A  man  engaged  to  write  for  20  days,  receiving  $2.50 
for  every  day  he  labored,  and  forfeiting  $1  for  every  day  he 
was  idle ;  at  the  end  of  the  time  he  received  $43  ;  how  many 
days  did  he  labor  ?  Ans.    18. 

47.  A,  B,  and  C  can  perform  a  piece  of  work  in  12  hours ; 
A  and  B  can  do  it  in  16  hours,  and  A  and  C  in  18  hours; 
what  part  of  the  work  can  B  and  C  do  in  9^  hours  ?    Ans.  $•. 


ALLIGATION  MEDIAL.  305 

ALLIGATION. 

1OO.  Alligation  treats  of  mixing  or  compounding  two  or 
more  ingredients  of  different  values.  It  is  of  two  kinds  —  Alii- 
gallon  Medial  and  Alligation  Alternate. 

4O1 .  Alligation  Medial  is  the  process  of  finding  the  aver- 
age  price  or  quality  of  a  compound  of  several  simple  ingredi- 
ents whose  prices  or  qualities  are  known. 

1.  A  miller  mixes   40  bushels  of  rye  worth  80  cents  a 
bushel,  and  25  bushels  of  corn  worth  70  cents  a  bushel,  with 
15  bushels  of  wheat  worth  $1.50  a  bushel ;  what  is  the  value 
of  a  bushel  of  the  mixture  ? 

OPERATION.  ANALYSIS.     Since  40  bushels 

80  X  40  =  $32.00  of  rye  at  80  cents  a  bushel  is 

70  X  25  =     1 7.50  worth  $32,  and  25  bushels  of  corn 

i  *;n  \x  i  ^  —  .    99  ^ft  at  70  cents   a  bushel  is   worth 

$17.50,  and  15  bushels  of  wheat 

80       )  72.00  at  $1.50  a  bushel  is  worth  $22.50, 

$90  Ans         therefore  the  entire  mixture,  con- 
sisting of  80  bushels,  is  worth 

$72,  and  one  bushel  is  worth  -^  of  $72,  or  72  -|-  80  =  $.90. 
Hence  the  following 

RULE.  Divide  the  entire  cost  or  value  of  the  ingredients 
by  the  sum  of  the  simples. 

EXAMPLES    FOR    PRACTICE. 

2.  A  wine  merchant  mixes  12  gallons  of  wine,  at   $1  per 
gallon,  with  5  gallons  of  brandy  worth  $1.50  per  gallon,  and 
3  gallons  of  water  of  no  value ;  what  is  the  worth  of  one  gal- 
lon of  the  mixture  ?  Ans.   $.975. 

3.  An  innkeeper  mixed  13  gallons  of  water  with  52  gallons 
of  brandy,  which  cost  him  $1.25  per  gallon  ;  what  is  the  value 
of  1  gallon  of  the  mixture,  and  what  his  profit  on  the  sale  of 
the  whole  at>  6£  cents  per  gill  ?    Ans.    $1  a  gallon ;  $65  profit. 

4.  A  grocer  mixed   10  pounds  of  sugar  at  8  cts.  with  12 
pounds  at  9  cts.  and  16  pounds  at  11  cts.,  and  sold  the  mixture 
at  10  cents  per  pound ;  did  he  gain  or  lose  by  the  sale,  and 
how  much  ?  Ans.   He  gained  1 6  cts. 

z* 


306  ALLIGATION   ALTERNATE. 

5.  A  grocer  bought  7£  dozen  of  eggs  at  12  cents  a  dozen, 
8  dozen  at  10£  cents  a  dozen,  9  dozen  at  11   cents  a  dozen, 
and  10£  dozen  at  10  cents  a  dozen.     He  sells  them  so  as  to 
make  50  per  cent,  on  the  cost  ;  how  much  did  he  receive  per 
dozen?  Am.    15£f  cents. 

6.  Bought  4  cheeses,  each  weighing  50  pounds,  at  13  cents 
a  pound;  10,  weighing  40  pounds  each,  at  10  cents  a  pound; 
and  24,  weighing  25   pounds  each,  at  7  cents  a  pound  ;  I  sold 
the  whole  at  an  average  price  of  9£  cents  a  pound  ;  how  much 
was  my  whole  gain  ?  Ans.    $6. 

4OS.  Alligation  Alternate  is  the  process  of  finding  the 
proportional  quantities  to  be  taken  of  several  ingredients, 
whose  prices  or  qualities  are  known,  to  form  a  mixture  of  a 
required  price  or  quality. 

CASE    I. 

4LO3.  To  find  the  proportional  quantity  to  be  used 
of  each  ingredient,  when  the  mean  price  or  quality  of 
the  mixture  is  given. 

1.  What  relative  quantities  of  timothy  seed  worth  $2  a 
bushel,  and  clover  seed  worth  $7  a  bushel,  must  be  used  to 
form  a  mixture  worth  $5  a  bushel  ? 

OPERATION.  ANALYSIS.    Since  on  every  in- 


2 


gredient  used  whose  price  or  qual- 

fr  is  less  tnan  tlie  mean  rate  tnere 


7 

will  be  a  gain,  and  on  every  in- 
gredient whose  price  or  quality  is  greater  than  the  mean  rate 
there  will  be  a  loss,  and  since  the  gains  and  losses  must  be  exactly- 
equal,  the  relative  quantities  used  of  each  should  be  such  as  repre- 
sent the  unit  of  value.  By  selling  one  bushel  of  timothy  seed  worth 
$2,  for  $o,  there  is  a  gain  of  $3  ;  and  to  gain  $1  would  require  ^  of 
a  bushel,  which  we  place  opposite  the  2.  By  selling  one  bushel  of 
clover  seed  worth  $7,  for  $5,  there  is  a  loss  of  $2  ;  and  to  lose  $1 
would  require  \  of  a  bushel,  which  we  place  opposite  tfie  7. 

In  every  case,  to  find  the  unit  of  value  we  must  divide  $1  by  the 
gain  or  loss  per  bushel  or  pound,  &c.  Hence,  if,  every  time  we  take 
•^  of  a  bushel  of  timothy  seed,  we  take  \  of  a  bushel  of  clover  seed, 
the  gain  and  loss  will  be  exactly  equal,  and  \ve  shall  have  \  and  \ 
for  the  proportional  quantities. 


ALLIGATION    ALTERNATE. 


307 


OPERATION. 


6 


1 

2 

3 

4 

5 

3 

i 

4 

4 

4 

i 

1 

1 

7 

1 

2 

2 

.  10 

i 

3 

3 

If  we  wish  to  express  the  proportional  numbers  in  integers,  we 
may  reduce  these  fractions  to  a  common  denominator,  and  use  their 
numerators,  since  fractions  having  a  common  denominator  are  to 
each  other  as  their  numerators.  (365)  thus,  £  and  £  are  equal  to 
|  and  f ,  and  the  proportional  quantities  are  2  bushels  of  timothy 
seed  to  3  bushels  of  clover  seed. 

2.  What  proportions  of  teas  worth  respectively  3,  4,  7  and 
10  shillings  a  pound,  must  be  taken  to  form  a  mixture  worth 
6  shillings  a  pound  ? 

ANALYSIS.  To  preserve  the 
equality  of  gains  and  losses,  we 
must  always  compare  two  prices 
or  simples,  one  greater  and  one 
less  than  the  mean  rate,  and 
treat  each  pair  or  couplet  as  a 
separate  example.  In  the  given 
example  we  form  two  couplets, 

and  may  compare  either  3  and  10,  4  and  7,  or  3  and  7,  4  and  10. 
We  find  that  £  of  a  Ib.  at  3  s.  must  be  taken  to  gain  1  shilling, 

and  \  of  a  Ib.  at  10  s.  to  lose  1  shilling  ;  also  \  of  a  Ib.  at  4  s.  to  gain 

1  shilling,  and  1  Ib.  at  7  s.  to  lose  1  shilling.     These  proportional 
numbers,  obtained  by  comparing  the  two  couplets,  are  placed  in 
columns  1  and  2.     If,  now,  we  reduce  the  numbers  in  columns  1  and 

2  to  a  common  denominator,  and  use  their  numerators,  we  obtain 
the  integral  numbers  in  columns  3  and  4,  which,  being  arranged  in 
column  5,  give  the  proportional  quantities  to  be  taken  of  each.* 

It  will  be  seen  that  in  comparing  the  simples  of  any  couplet,  one 
of  which  is  greater,  and  the  other  less  than  the  mean  rate,  the  pro- 
portional number  finally  obtained  for  either  term  is  the  difference 
between  the  mean  rate  and  the  other  term.  Thus,  in  comparing  3 
and  10,  the  proportional  number  of  the  former  is  4,  which  is  the 
difference  between  10  and  the  mean  rate  6  ;  and  the  proportional 
number  of  the  latter  is  3,  which  is  the  difference  between  3  and  the 
mean  rate.  The  same  is  true  of  every  other  couplet.  Hence,  when 
the  simples  and  the  mean  rate  are  integers,  the  intermediate  steps 
taken  to  obtain  the  final  proportional  numbers  as  in  columns  1,  2,  3, 
and  4,  may  be  omitted,  and  the  same  results  readily  found  by  taking 
the  difference  between  each  simple  and  the  mean  rate,  and  placing 
it  opposite  the  one  with  which  it  is  compared. 

*  Prof.  A.  B.  Canfield,  of  Oneida  Conference  Seminary,  N.  Y.,  used  this  method  of 
Alligation,  essentially,  in  the  instruction  of  his  classes  as  early  as  1846,  and  he  was 
doubtless  the  author  of  it. 


808  ALLIGATION  ALTERNATE. 

From  the  foregoing  examples  and  analyses  we  derive  the  following 

RULE.     I.    Write  the  several  prices  or  qualities  in  a  column, 
and  the  mean  price  or  quality  of  the  mixture  at  the  left. 

II.  Form  couplets  ly  comparing  any  price  or  quality  Iess9 
with  one  that  is  greater  than  the  mean  rate,  placing  the  part 
which  must  be  used  to  gain  1  of  the  mean  rate  opposite  the  less 
simple,  and  the  part  that  must  be  used  to  lose  1  opposite  the 
greater  simple,  and  do  the  same  for  each  simple  in  every  couplet. 

III.  If  the  proportional  numbers  are  fractional,  they  may  be 
reduced  to  integers,  and  if  two  or  more  stand  in  the  same  hori- 
zontal line,  they  must  be  added;  the  jinal  results  will  be  the  pro- 
portional quantities  required. 

NOTES.     1.  If  the  numbers  in  any  couplet  or  column  have  a  com- 
mon factor,  it  may  be  rejected. 

2.  We  may  also  multiply  the  numbers  in  any  couplet  or  column  by 
any  multiplier  we  choose,  without  affecting  the  equality  of  the  gains 
and  losses,  and  thus  obtain  an  indefinite  number  of  results,  any  one  of 
which  being  taken  will  give  a  correct  final  result. 

EXAMPLES   FOR  PRACTICE. 

3.  A  grocer  has  sugars  worth  10  cents,  11  cents,  and  14 
cents  per  pound ;  in  what  proportions  may  he  mix  them  to 
form  a  mixture  worth  12  cents  per  pound? 

Ans.   2  Ibs.  at  10  and  11  cts.,  and  3  Ibs.  at  14cts. 

4.  What  proportions  of  water  at  no  value,  and  wine  worth 
$1.20  a  gallon,  must  be  used  to  form,  a  mixture  worth  90  cents 
a  gallon  ?  Ans.    1  gal.  of  water  to  3  gals,  of  wine. 

5.  A  farmer  had  sheep  worth  $2,  $2£,  $3,  and  $4  per 
head ;  what  is  the  smallest  number  he  could  sell  of  each,  and 
realize  an  average  price  of  $2f  per  head  ? 

Ans.   3  of  the  1st  kind,  1  each  of  the  2d  and  3d,  and  5  of 
the  4th  kind. 

6.  What  relative  quantities  of  alcohol  80,  84,  87,  94,  and 
96  per  cent,  strong  must  be  used  to  form  a  mixture  90  per 
cent,  strong  ? 

Ans.   6  of  the  first  two  kinds,  four  of  the  3d,  3  of  the  4th, 
and  16  of  the  5th. 


ALLIGATION  ALTERNATE.  309 

CASE  II. 

4O4.  When  the  quantity  of  one  of  the  simples  is 
limited. 

1.  A  miller  has  oats  worth  30  cents,  corn  worth  45  cents, 
and  barley  worth  84  cents  per  bushel ;  he  desires  to  form  a 
mixture  worth  60  cents  per  bushel,  and  which  shall  contain  40 
bushels  of  corn ;  how  many  bushels  of  oats  and  barley  must 
he  take  ? 

OPERATION.  ANALYSIS.  By 


S30 
45 
84 


4 

8 
5  |  5 


20   )  the  same  process 

40   (  Ans.      as  in  Case  l  we 
50  1  fi.nd  the  Pro?or- 


tional  quantities 
of  each  to,  be  4  bushels  of  oats,  8  of  corn,  and  10  of  barley.  But 
we  wish  to  use  40  bushels  of  corn,  which  is  5  times  the  propor- 
tional number  8,  and  to  preserve  the  equality  of  gain  and  loss  we 
must  take  5  times  the  proportional  quantity  of  each  of  the  other 
simples,  or  5  X  4  =  20  bushels  of  oats,  and  5  X  10  =  50  bushels 
of  barley.  Hence  the  following 

RULE.  Find  the  proportional  quantities  as  in  Case  L 
Divide  the  given  quantity  by  the  proportional  quantity  of  the 
same  ingredient,  and  multiply  each  of  the  other  proportional 
quantities  by  the  quotient  thus  obtained. 

EXAMPLES    FOR    PRACTICE. 

2.  A  merchant  has  teas  worth  40,  60,  75,  and  90  cents  per 
pound  ;  how  many  pounds  of  each  must  he  use  with  20  pounds 
at  75  cents,  to  form  a  mixture  worth  80  cents  ? 

Ans.  201bs.  each  of  the  first  three  kinds,  and  1301bs.  of 
the  fourth. 

3.  A  farmer  bought  24  sheep  at  $2  a  head ;  how  many 
must  he  buy  at  $3  and  $5  a  head,  that  he  may  sell  the  whole 
at  an  average  price  of  $4  a  head,  without  loss  ? 

Ans.   24  at  $3,  and  72  at  $5. 

4.  How  much  alcohol  worth  60  cents  a  gallon,  and  how 
much  water,  must  be  mixed  with  180  gallons  of  rum  worth 
$1.30  a  gallon,  that  the  mixture  may  be  worth  90  cents  a 
gallon  ?  Ans.   60  gallons  each  of  alcohol  and  of  water. 


310 


ALLIGATION    ALTERNATE. 


5.  How  many  acres  of  land  worth  35  dollars  an  acre  must 
be  added  to  a  farm  of  75  acres,  worth  $50  an  acre,  that  the 
average  value  may  be  $40  an  acre  ?  Am.    150  acres. 

6.  A  merchant  mixed  80  pounds  of  sugar  worth  6J  cents 
per  pound  with  some  worth  8£  cents  and  10  cents  per  pound, 
so  that  the  mixture  was  worth  7£  cents  per  pound  ;  how  much 
of  each  kind  did  he  use  ? 

CASE    III. 

4:0*5.  When  the  quantity  of  the  whole  compound  is 
limited. 

1.  A  grocer  has  sugars  worth  6  cents,  8  cents,  11  cents, 
and  13  cents  per  pound.     He  wishes  to  make  a  mixture  of 
120  pounds  worth  10  cents  a  pound;  how  many  pounds  of 
each  kind  must  he  use  ? 

ANALYSIS.  By  Case 
I  we  find  the  propor- 
tional  quantities  of  each 
to  be  3  Ibs.  at  6  cts.,  2 
Ibs.  at  7  cts.,  3  Ibs  at  12 
cts.,  and  4  Ibs.  at  13  cts. 
By  adding  the  propor- 
tional quantities,  we  find 
that  the  mixture  would  be  but  12  Ibs.  while  the  required  mixture  is 
120,  or  10  times  12.  If  the  whole  mixture  is  to  be  10  times  as  much 
as  the  sum  of  the  proportional  quantities,  then  the  quantity  of  each 
simple  used  must  be  10  times  as  much  as  its  respective  proportion- 
al, which  would  require  30  Ibs.  at  6  cts.,  20  Ibs.  at  7  cts.,  30  Ibs.  at 
12  cts.,  and  40  Ibs.  at  13  cts.  Hence  we  deduce  the  following 

RULE.  Find  the  proportional  numbers  as  in  Case  I.  Di- 
vide the  given  quantity  by  the  sum  of  the  proportional  quan- 
tities, and  multiply  each  of  the  proportional  quantities  by  the 
quotient  thus  obtained. 

EXAMPLES    FOR    PRACTICE. 

2.  A  farmer  sold   170  sheep  at  an  average  price  of  14 
shillings  a  head  ;  for  some  he  received  9  s.,  for  some  12  s.,  for 
some  18s.,  and  for  others  20s.;  how  many  of  each  did  he 
sell  ?     Ans.   60  at  9  s.,  40  at  12  s.,  20  at  18  s.,  and  50  at  20  s. 


OPERATION. 

6 

i 

3 

3 

30 

7 

% 

2 

2 

20 

12 

% 

3 

3 

30 

13 

i 

4 

4 

40 

12 

120 

INVOLUTION.  311 

3.  A  jeweler   melted   together  gold  16,  18,  21,  and    24 
carats  fine,  so  as  to  make  a  compound  of  51  ounces  22  carats 
fine ;  how  much  of  each  sort  did  he  take  ?     Ans.    6  ounces 
each  of  the  first  three,  and  33  ounces  of  the  last. 

4.  A  man  bought  210  bushels  of  oats,  corn,  and  wheat,  and 
paid  for  the  whole  $178.50  ;    for  the  oats  he  paid  $£,  for  the 
corn  $-£,  and  for  the  wheat  $1£  per  bushel ;  how  many  bush- 
els of  each  kind  did  he  buy  ?     Ans.    78  bushels  each  of  oats 
and  corn,  and  54  bushels  of  wheat. 

5.  A,  B,  and  C  are  under  a  joint  contract  to  furnish  6000 
bushels  of  corn,  at  48  cts.  a  bushel ;  A's  corn  is  worth  45  cts., 
B's  51  cts.,  and  C's  54  cts. ;  how  many  bushels  must  each  put 
into  the  mixture  that  the  contract  may  be  fulfilled  ? 

6.  One  man  and  3  boys  received  $84  for  56  days'  labor;  the 
man  received  $3  per  day,  and  the  boys  $£,  $},  and  $lf  re- 
spectively ;  how  many  days  did  each  labor  ?     Ans.   The  man 
16  days,  and  the  boys  24,  4,  and  12  days  respectively. 

INVOLUTION. 

406.  A  Power  is  the  product  arising  from  multiplying  a 
number  by  itself,  or  repeating  it  several  times  as  a  factor; 
thus,  in  2  X  2  X  2  =  8,  the  product,  8,  is  a  power  of  2. 

407.  The  Exponent  of  a  power  is  the  number  denoting 
how  many  times  the  factor  is  repeated  to  produce  the  power, 
and  is  written  above  and  a  little  to  the  right  of  the  factor;  thus, 
2  X  2  X  2  is  written  23,  in  which  3  is  the  exponent.    Exponents 
likewise  give  names  to  the  powers,  as  will  be  seen  in  the 
following  illustrations : 

3  =  3l  —    3,  the  first  power  of  3 ; 

3X3  =  32  =    9,  the  second  power  of  3 

3X3X3  =  33  ==.  27,  the  third  power  of  3. 

408.  The  Square  of  a  number  is  its  second  power. 
4:09.    The  Cube  of  a  number  is  its  third  power. 

41O.    Involution  is  the  process  of  raising  a  number  to  a 
given  power. 


312  EVOLUTION. 

411.  A  Perfect  Power  is  a  number  that  can  be  exactly 
produced  by  the  involution  of  some  number  as  a  root ;  thus,  25 
and  32  are  perfect  powers,  since  25  =  5  X  5,  and  32  =  2  X 
2X2X2X2. 

I.  What  is  the  cube  of  15  ? 

OPERATION.  ANALYSIS.    We  multiply 

15  X  15  X  15  =  3375.   Ans.        l5  b?  15'  and  the  product 

by   15,   and    obtain    3375, 

•which  is  the  3d  power,  or  cube  of  15,  since  15  has  been  taken  3 
times  as  a  factor.     Hence,  we  have  the  following 

RULE.    Multiply  the  number  by  itself  as  many  times,  less  1, 
as  there  are  units  in  the  exponent  of  the  required  power. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  square  of  25  ?  Ans.    625. 

3.  What  is  the  square  of  135  ?  Ans.    18225. 

4.  What  is  the  cube  of  72  ?  Ans.   373248. 

5.  What  is  the  4th  power  of  24?  Ans.   331776. 

6.  Raise  7.2  to  the  third  power.  Ans.   373.248. 

7.  Involve  1.06  to  the  4th  power.  Ans.    1.26247696. 

8.  Involve  .12  to  the  5th  power.  Ans.   .0000248832. 

9.  Involve  1.0002  to  the  2d  power.  Ans.  1.00040004. 
10.  What  is  the  cube  off? 

OPERATION. 

JL      —       2  ^2X2X2       2*        8 
T  X  5  X  5  ~"  5~X~5X  5  ~~  53~~  125 

It  is  evident  from  the  above  operation,  that 

A  common  fraction  may  be  raised  to  any  power,  by  raising 
each  of  its  terms,  separately,  to  the  required  power. 

II.  What  is  the  square  of  f  ?  Ans.   ¥9¥. 

12.  What  is  the  cube  of  |f  ?  Ans.   f  f  f }. 

13.  Raise  24J  to  the  2d  power.  Ans.    612T%. 

EVOLUTION. 

412.  A  Boot  is  a  factor  repeated  to  produce  a  power ; 
thus,  in  the  expression  5  X  5  X  5  =  125,  5  is  the  root  from 
which  the  power,  125,  is  produced. 


SQUARE   HOOT.  313 

S  I  J{.  Evolution  is  the  process  of  extracting  the  root  of  a 
number  considered  as  a  power,  and  is  the  reverse  of  Involution. 

414.  The  Radical  Sign  is  the  character,  V>  which,  placed 
before  a  number,  denotes  that  its  root  is  to  be  extracted. 

41t>.  The  Index  of  the  root  is  the  figure  placed  above  the 
radical  sign,  to  denote  what  root  is  to  be  taken.  When  no 
index  is  written,  the  index  2  is  always  understood. 

416.  A  Surd  is  the  indicated  root  of  an  imperfect  power. 

417.  Roots  are  named  from  the  corresponding  powers,  as 
will  be  seen  in  the  following  illustrations : 

The  square  root  of  9  is  3,  written  -\/9  i=  3. 
The  cube  root  of  27  is  3,  written  ^27  =  3. 
The  fourth  root  of  81  is  3,  written  ^81  =  3. 

418.  Any  number  whatever  may  be  considered  a  power 
whose  root  is  to  be  extracted  ;  but  only  the  perfect  powers  can 
have  exact  roots. 

SQUARE  ROOT. 

419.  The  Square  Root  of  a  number  is  one  of  the  two 

equal  factors  that  produce  the  number ;  thus  the  square  root 

of  49  is  7,  for  7  X  7  =  49. 

420.  In  extracting  the  square  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  given  number 
and  its  square  root.    The  law  governing  -this  relation  is  exhib- 
ited in  the  following  examples  :  — 

Roots.  Squares.  Roots.  Squares. 

Ill  1 

9  81  10  1,00 

99  98,01  100  1,00,00 

999  99,80,01  1000  1,00,00,00 

From  these  examples  we  perceive 

1st.  That  a  root  consisting  of  1  place  may  have  1  or  2 
places  in  the  square. 

2d.    That  in  all  cases  the  addition  of  1   place  to  the  root 
adds  2  places  to  the  square.     Hence, 
A  A 


314  SQUARE   ROOT. 

If  we  point  off  a  number  into  two-figure  periods,  commen- 
cing at  the  right  hand,  the  number  of  full  periods  and  the  left 
hand  full  or  partial  period  will  indicate  the  number  of  places 
in  the  square  root ;  the  highest  period  answering  to  the  highest 
figure  of  the  root. 

421.  1.  What  is  the  length  of  one  side  of  a  square  plat 
containing  an  area  of  5417  sq.  ft.  ? 

OPERATION.  ANALYSIS.     Since  the  given  figure  is 

54,17  |  73.6  a  square,  its  side  will  be  the  square  root 

49  of  its  area,  which  we  will  proceed  to  com- 

pute.   Pointing  off  the  given  number,  the 

2  periods  show  that  there  will  be  two  in- 
tegral figures,  tens  and  units,  in  the  root. 

146.0     88.00  ^e  tens  °f  ^e  root  must  be  extracted 

1 46  6     87  96  from  the  first  or  left  hand  period,  54  hun- 

dreds.    The  greatest  square  in  54  hun- 
4  dreds  is  49  hundreds,  the  square  of  7  tens  ; 

we  therefore  write  7  tens  in  the  root,  at 
the  right  of  the  given  number. 

Since  the  entire  root  is  to  be  the  side  of  a  square,  let  us  form  a 
Fig.  i.  square  (Fig.  I),  the  side  of  which  is  70  feet  long. 

The  area  of  this  square  is  70  X  70  =  4900  sq.ft., 
which  we  subtract  from  the  given  number.  This 
is  done  in  the  operation  by  subtracting  the 
square  number,  49,  from  the  first  period,  54, 
and  to  the  remainder  bringing  down  the  sec- 
ond period,  making  the  entire  remainder  517. 
If  we  now  enlarge  our  square  (Fig.  I)  by  the  addition  of  517 
square  feet,  in  such  a  manner  as  to  preserve  the  square  form,  its 
size  will  be  that  of  the  required  square.  To  preserve  the  square 
form,  the  addition  must  be  so  made  as  to  extend  the  square  equally 
in  two  directions  ;  it  will  therefore  be  composed  of  2  oblong  figures 
at  the  sides,  and  a  little  square  at  the  corner  (Fig.  II).  Now,  the 
width  of  this  addition  will  be  the  additional  length  to  the  side  of  the 
square,  and  consequently  the  next  figure  in  the  root.  To  find  width 
we  divide  square  contents,  or  area,  by  length.  But  the  length  of 
one  side  of  the  little  square  cannot  be  found  till  the  width  of  the 
addition  be  determined,  because  it  is  equal  to  this  width.  We  will 
therefore  add  the  lengths  of  the  2  oblong  figures,  and  the  sum  will 
be  sufficiently  near  the  whole  length  to  be  used  as  a  trial  divisor. 


SQUARE  ROOT. 


315 


Fig.  H. 


Fig.  III. 


70 

70. 

3 

i 
;              Two  Divisors  =  140 

Complete  Divisor  =  143 

Each  of  the  oblong  figures  is  equal  in  length  to  the  side  of  the 
square  first  formed  ;  and  their  united  length 
is  70  +  70  =  140  ft.  (Fig.  III).  This  num- 
ber is  obtained  in  the  operation  by  doubling 
the  7  and  annexing  1  cipher,  the  result  being 
written  at  the  left  of  the  dividend.  Dividing 
517,  the  area,  by  140,  the  approximate  length, 
we  obtain  3,  the  probable  width  of  the  addi- 
tion, and  second  figure  of  the  root.  Since  3  is 
also  the  side  of  the  little  square,  we  can  now 
find  the  entire  length  of  the  addition,  or  the  complete  divisor,  which 

is  70  +  70  -f  3  =  143  (Fig.  III). 
This  number  is  found  in  the  oper- 
ation by  adding  3  to  the  trial  di- 
visor, and  writing  the  result  un- 
derneath. Multiplying  the  com- 
plete divisor,  143,  by  the  trial 
quotient  figure,  3,  and  subtracting 
the  product  from  the  dividend,  we 
obtain  another  remainder  of  88  square  feet.  With  this  remainder, 
for  the  same  reason  as  before,  we  must  proceed  to  make  a  new 
enlargement ;  and  we  bring  down  two  decimal  ciphers,  because  the 
next  figure  of  the  root,  being  tenths,  its  square  will  be  hundredths. 
The  trial  divisor  to  obtain  the  width  of  this  new  enlargement, 
or  the  next  figure  in  the  root,  will  be,  for  the  same  reason  as 
before,  twice  73,  the  root  already  found,  with  one  cipher  annexed. 
But  since  the  7  has  already  been  doubled  in  the  operation,  we  have 
only  to  double  the  last  figure  of  the  complete  divisor,  143,  and 
annex  a  cipher,  to  obtain  the  new  trial  divisor,  146.0.  Dividing,  we 
obtain  .6  for  the  trial  figure  of  the  root ;  then  proceeding  as  before, 
we  obtain  146.'6  for  a  complete  divisor,  87.96  for  a  product ;  and 
there  is  still  a  remainder  of  .04.  Hence,  the  side  of  the  given 
square  plat  is  73.6  feet,  nearly.  From  this  example  and  analysis 
we  deduce  the  following 

RULE.  I.  Point  off  the  given  number  into  periods  of  two 
figures  each,  counting  from  unit's  place  toward  the  left  and 
right. 

II.  Find  the  greatest  square  number  in  the  left  hand  period, 
and  write  its  root  for  the  first  figure  in  the  root ;  subtract  the 
square  number  from  the  left  hand  period,  and  to  the  remainder 
bring  down  the  next  period  for  a  dividend. 


316  SQUARE   HOOT. 

III.  At  the  left  of  the  dividend  write  twice  the  first  figure  of 
the  root,  and  annex  one  cipher,  for  a  trial  divisor  ;  divide  the 
dividend  by  the  trial  divisor,  and  write  the  quotient  for  a  trial 
figure  in  the  root. 

IV.  Add  the  trial  figure  of  the  root  to  the  trial  divisor  for  a 
complete  divisor ;  multiply  the  complete   divisor  by  the  trial 
figure  in  the  root,  and  subtract  the  product  from  the  dividend, 
and  to  the  remainder  bring  down  the  next  period  for  a  new 
dividend. 

V.  Multiply  the  last  figure  of  the  last  complete  divisor  by  2, 
and  annex  one  cipher  for  a  new  trial  divisor,  with  which  pro- 
ceed as  before. 

NOTES.  1.  If  at  any  time  the  product  be  greater  than  the  dividend, 
diminish  the  trial  figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  another  cipher  to  the  trial 
divisor,  and  another  period  to  the  dividend,  and  proceed  as  before. 

EXAMPLES    FOR   PRACTICE. 

2.   What  is  the  square  root  of  406457.2516? 

OPERATION. 

40,64,57.25,16    637.54,  Ans. 
36 


Trial    divisor, 
Complete  " 

Trial         " 
Complete  " 

Trial         « 
Complete  " 

Trial          " 
Complete  " 

120 
123 

464 
369 

1260 

1267 

9557 
8869 

1274.0 
1274.5 

688.25 
637.25 

1275.00 
1275.04 

51.0016 
'    51.0016 

NOTES.  3.  The  decimal  points  in  the  work  may  be  omitted,  care 
being  taken  to  point  off  in  the  root  according  to  the  number  of  deci- 
mal periods  used. 

4.  The  pupil  will  acquire  greater  facility,  and  secure  greater  accura- 
cy, by  keeping  units  of  like  order  under  each  other,  and  each  divisor 
opposite  the  corresponding  dividend,  by  the  use  of  the  lines,  as  shown 
in  the  operation. 

3.   What  is  the  square  root  of  576  ?  Ans.    24. 


SQUARE  ROOT.  317 

4.  What  is  the  square  root  of  6561  ?  Am.    81. 

5.  What  is  the  square  root  of  444889  ?  Ans.    667. 

6.  What  is  the  square  root  of  994009  ?  Ans.    997. 

7.  What  is  the  square  root  of  29855296?        Ans.   5464. 

8.  What  is  the  square  root  of  3486784401  ?    Ans.  59049. 

9.  What  is  the  square  root  of  54819198225  ? 

NOTE.  The  cipher  in  the  trial  divisor  may  be  omitted,  and  its  place, 
after  division,  occupied  by  the  trial  root  figure,  thus  forming  in  suc- 
cession only  complete  divisors. 

10.   What  is  the  square  root  of  2  ? 

2.         |  1.4142  +,  Ans. 


100 
24         96 

400 
281         281 

11900 
2824       11296 

60400 
28282         56564 

11.   Extract  the  square  roots  of  the  following  numbers: 


V3  =  1.7320508  + 
V5  =  2.2360680 -I'- 
ve =  2.4494897  + 


V7  =  2.6457513  + 
V8  =  2.8284271  + 
V10  =  3.1622776  + 


12.  What  is  the  square  root  of  .00008836  ?       Ans.   .0094. 

13.  What  is  the  square  root  of  .0043046721  ?  Ans.  .06561. 

NOTES.  5.  The  square  root  of  a  common  fraction  may  be  obtained 
by  extracting  the  square  roots  of  the  numerator  and  denominator 
separately,  provided  the  terms  are  perfect  squares ;  otherwise,  the 
fraction  may  first  be  reduced  to  a  decimal. 

6.  Mixed  numbers  may  be  reduced  to  the  decimal  form  before  ex- 
tracting the  root ;  or,  if  the  denominator  of  the  fraction  is  a  perfect 
square,  to  an  improper  fraction. 

14.  Extract  the  square  root  of  ^5T.  Ans.   ff. 

15.  Extract  the  square  root  of  ££^f.  Ans.   £. 

16.  Extract  the  square  root  of  §.      ,  Ans.    .81649  +  . 

17.  Extract  the  square  root  of  17f.  Ans.    4.168  +  . 

AA* 


318 


SQUARE  ROOT 


APPLICATIONS. 

2.  An  Angle  is  the  opening  between  two  lines  that 
meet  each  other ;  thus,  the  two  lines,  A  B  and  A  C,  meeting, 
form  an  angle  at  A. 

4123.  A  Triangle  is  a  figure  having  three 
sides  and  three  angles,  as  A,  B,  C. 

424.  A  Right-Angled  Triangle  is  a  tri- 
angle having  one  right  angle,  as  at  C. 

425.  The  Base  is  the  side  on  which  it 
stands,  as  A,  C. 

4:26.    The  Perpendicular  is  the   side 
forming  a  right  angle  with  the  base,  as  B,  C. 

427.  The  Hypotenuse  is  the  side  opposite  the  right  angle, 
as  A,  B. 

4:28.  Those  examples  given  below,  which  relate  to  trian- 
gles and  circles,  may  be  solved  by  the  use  of  the  two  following 
principles,  which  are  demonstrated  in  geometry. 

1st.  The  square  of  the  hypotenuse  of  a  right-angled  triangle 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides. 

2d.  The  areas  of  two  circles  are  to  each  other  as  the  squares 
of  their  radii,  diameters,  or  circumferences. 

1.  The  two  sides  of  a  right-angled  triangle  are  3  and  4 
feet ;  what  is  the  length  of  the  hypotenuse  ? 

ANALYSIS.     Squaring 

OPERATION.  the  two  sides  and  add- 

32  =    9,  square  of  one  side.  ing»  we  find  the  sum  to 

42  =  1 6,  square  of  the  other  side-       ?e  25  >  ,and  sjnce  the  sum, 

is  equal  to  the  square  of 

25,  square  of  hypotenuse.  the  hypotenuse,  \ve  cx- 

/25  _  5    ^nSt  tract  the  square  root,  and 

'  obtain  5  feet,  the  hypot- 

enuse.    Hence, 

To  find  the  hypotenuse.  Add  the  squares  of  the  two  sides, 
and  extract  the  square  root  of  the  sum. 

To  find  cither  of  the  shorter  sides.  Subtract  the  square  of 
flic  f liven  side  from  the  square  of  the  hypotenuse,  and  extract  the 
square  root  of  the  remainder. 


SQUARE  ROOT.  319 

EXAMPLES   FOR    PRACTICE. 

2.  If  an  army  of  55225  men  be  drawn  up  in  the  form  of  a 
square,  how  many  men  will  there  be  on  a  side  ?    Ans.   235. 

3.  A  man  has  200  yards  of  carpeting  1£  yards  wide  ;  what 
is  the  length  of  one  side  of  the  square  room  which  this  carpet 
will  cover  ?  Ans.    45  feet. 

4.  How  many  rods  of  fence  will  be  required  to  inclose  10 
acres  of  land  in  the  form  of  a  square  ?  Ans.    1 60  rods. 

5.  The  top  of  a  castle  is  45  yards  high,  and  the  castle  is  sur- 
rounded by  a  ditch  60  yards  wide ;  required  the  length  of  a 
rope  that  will  reach  from  the  outside  of  the  ditch  to  the  top 
of  the  castle.  Ans.    75  yards. 

6.  Required  the  height  of  a  May-pole,  which  being  broken 
39  feet  from  the  top,  the  end  struck  the  ground  15  feet  from 
the  foot.  Ans.   75  feet. 

7.  A  ladder   40    feet  long   is  so  placed  in  a  street,  that 
without  being  moved  at  the  foot,  it  will  reach  a  window  on 
one  side  33  feet,  and  on  the  other  side  21   feet,  from  the 
ground  ;  what  is  the  breadth  of  the  street  ?     Ans.  56.65  -f-  ft. 

8.  A  ladder  52  feet  long  stands  close  against  the  side  of  a 
building  ;  how  many  feet  must  it  be  drawn  out  at  the  bottom, 
that  the  top  may  be  lowered  4  feet  ?  Ans.    20  feet. 

9.  Two  men  start  from  one  corner   of  a  park  one  mile 
square,  and  travel  at  the  same  rate.     A  goes  by  the  walk 
around  the  park,  and  B  takes  the  diagonal  path  to  the  opposite 
corner,  and  turns  to  meet  B  at  the  side.     How  many  rods 
from  the  corner  will  the  meeting  take  place  ?    Ans.  93.7  -\-  rods. 

10.  A  room  is  20  feet  long,  16  feet  wide,  and  12  feet  high ; 
what  is  the  distance  from  one  of  the  lower  corners  to  the  op- 
posite upper  corner  ?  Ans.   28.284271  -f-  feet. 

11.  It  requires  63.39  rods  of  fence  to  inclose  a  circular 
field  of  2  acres ;  what  length  will  be  required  to  inclose  3 
acres  in  circular  form  ?  Ans.   77.63  rods. 

12.  The, radius  of  a  certain  circle  is  5  feet ;  what  will  be 
the  radius  of  another  circle  containing  twice  the  area  of  the 
first?  Ans.   7.07106  + feet. 


820  CUBE  ROOT. 


CUBE   ROOT. 

4:29.  The  Cube  Koot  of  a  number  is  one  of  the  three 
equal  factors  that  produce  the  number.  Thus,  the  cube  root 
of  27  is  3,  since  3  X  3  X  3  =  27. 

4:30  »  In  extracting  the  cube  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  cube  and  its 
root.  The  law  governing  this  relation  is  exhibited  in  the  fol- 
lowing examples  :  — 

Roots.  Cubes.  Roots.  Cubes. 

Ill  1 

9  729  10  1,000 

99  907,299  100  1,000,000 

999         997,002,999  1000         1,000,000,000 

From  these  examples,  we  perceive, 

1st.  That  a  root  consisting  of  1  place  may  have  from  1  to 
3  places  in  the  cube. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root 
adds  three  places  to  the  cube.  Hence, 

If  we  point  off  a  number  into  three-figure  periods,  com- 
mencing at  the  right  hand,  the  number  of  full  periods  and  the 
left  hand  full  or  partial  period  will  indicate  the  number  of 
places  in  the  cube  root,  the  highest  period  answering  to  the 
highest  figure  of  the  root. 

4131.  1.  What  is  the  length  of  one  side  of  a  cubical  block 
containing  413494  solid  inches  ? 

OPERATION  —  COMMENCED.  ANALYSIS.  Since  the  block  is  a 
413494  |  74  cube,  its  side  will  be  the  cube  root  of 
343  its  solid  contents,  which  we  will  pro- 

comPute'     PointinS   off   the 


14700  70494.        T 

given  number,  the  two  periods  show 

that  there  will  be  two  figures,  tens  and 

units,  in  the  root.  The  tens  of  the  root  must  be  extracted  from  the 
first  period,  413  thousands.  The  greatest  cube  in  413  thousands  is 
343  thousands,  the  cube  of  7  tens  ;  we  therefore  write  7  tens  in  the 
root  at  the  right  of  the  given  number. 


CUBE   ROOT. 


321 


Since  the  entire  root  is  to  be  the  side  of  a  cube,  let  us  form  a 
Fig%  L  cubical  block  (Fig.  I),  the  side 

of  which  is  70  inches  in  length. 
The  contents  of  this  cube  are 
70  X  70  X  70 1=  343,000  solid 
inches,  which  we  subtract  from 
the  given  number.  This  is  done 
in  the  operation  by  subtracting 
the  cube  number,  343,  from  the 
first  period,  413,  and  to  the  re- 
mainder bringing  down  the  sec- 
ond period,  making  the  entire 
remainder  70494. 

If  we  now  enlarge  our  cubical 
block,  (Fig.  I),  by  the  addition  of  70494  solid  inches,  in  such  a 
manner  as  to  preserve  the  cubical  form,  its  size  will  be  that  of 
the  required  block.  To  preserve  the  cubical  form,  the  addition 
must  be  made  upon  three  adjacent  sides  or  faces.  The  addition 
will  therefore  be  composed  of  3  flat  blocks  to  cover  the  3  faces, 
(Fig.  II) ;  3  oblong  blocks  to  fill  the  vacancies  at  the  edges, 
(Fig.  Ill)  ;  and  1  small  cubical  block  to  fill  the  vacancy  at  the  cor- 
ner, (Fig.  IV).  Now,  the  thickness  of  this  enlargement  will  be  the 
additional  length  of  the  side  of  the  cube,  and,  consequently,  the 
second  figure  in  the  root.  To  find  thickness,  we  may  divide  solid 
Fig- II-  contents  by  surface,  or  area. 

But  the  area  of  the  3  oblong 
blocks  and  little  cube  cannot 
be  found  till  the  thickness  of 
the  addition  be  determined,  be- 
cause their  common  breadth  is 
equal  to  this  thickness.  We  will 
therefore  find  the  area  of  the 
three  flat  blocks,  which  is  suffi- 
ciently near  the  whole  area  to  be 
used  as  a  trial  divisor.  As  these 
are  each  equal  in  length  and 
breadth  to  the  side  of  the  cube 
whose  faces  they  cover,  the  whole 
area  of  the  three  is  70  X  70  X 
3  rr:  14700  square  inches.  This  number  is  obtained  in  the  operation 
by  annexing  2  ciphers  to  three  times  the  square  of  7  ;  the  result 
being  written  at  the  left  hand  of  the  dividend.  Dividing,  we  obtain 


322 


CUBE  ROOT. 


4,  the  probable  thickness  of  the  addition,  and  second  figure  of  the 
Fig  IIL  root.    With  this  assumed  figure, 

we  will  complete  our  divisor  by 
adding  the  area  of  the  4  blocks, 
before  undetermined.  The  3  ob- 
long blocks  are  each  70  inches 
long  ;  and  the  little  cube,  being 
equal  in  each  of  its  dimensions 
to  the  thickness  of  the  addition, 
must  be  4  inches  long.  Hence, 
their  united  length  is  70  -j-  70 
+  70  +  4  —  214.  This  number 
is  obtained  in  the  operation  by 
multiplying  the  7  by  3,  and  an- 
nexing the  4  to  the  product,  the 
result  being  written  in  column 
I,  on  the  next  line  below  the 
trial  divisor.  Multiplying  214, 
the  length,  by  4,  the  common 
width,  we  obtain  856,  the  area  of 
the  four  blocks,  which  added  to 
14700,  the  trial  divisor,  makes 
15556,  the  complete  divisor ;  and 
multiplying  this  by  4,  the  second 
figure  in  the  root,  and  subtract- 
ing the  product  from  the  divi- 
dend, we  obtain  a  remainder  of 
8270  solid  inches.  With  this  re- 
mainder, for  the  same  reason  as 
before,  we  must  proceed  to  make 
a  new  enlargement.  But  since 
we  have  already  two  figures  in 
the  root,  answering  to  the  two 
periods  of  the  given  number, 
the  next  figure  of  the  root  must 
be  a  decimal ;  and  we  therefore 
annex  to  the  remainder  a  period 
of  three  decimal  ciphers,  mak- 
ing 8270.000  for  a  new  dividend. 
The  trial  divisor  to  obtain  the 
thickness  of  this  second -enhir^e- 
ment,  or  the  next  figure  of  the  root,  will  be  the  area  of  three  new  flat 
blocks  to  cover  the  three  sides  of  the  cube  already  formed  j  and  this 


OPERATION  —  CONTINUED. 

413494  |  74 

I.  IT. 

343 

214  856 

14700  70494 
15556  62224 

8270.000 

Fig.  IV. 


CUBE  ROOT. 


323 


surface,  (Fig.  IV,)  is  composed  of  1  face  of  each  of  the  flat  blocks 
already  used,  2  faces  of  each  of  the  oblong  blocks,  and  3  faces  of 
the  little  cube.  But  we  have  in  the  complete  divisor,  15556,  1 
face  of  each  of  the  flat  blocks,  oblong  blocks,  and  little  cube ; 
and  in  the  correction  of  the  trial  divisor,  856,  1  face  of  each  of 
the  oblong  blocks  and  of  the  little  cube;  and  in  the  square  of 
the  last  root  figure,  16,  a  third  face  of  the  little  cube.  Hence,  16 
_|_  856  -}-  15556  =  16428,  the  significant  figures  of  the  new  trial 

divisor.    This 


OPERATION  —  CONTIN 
I.                      II. 

UED. 

413494|  74.5 

number  is  ob- 
tained in  the 
operation    by 
adding       the 
square  of  the 
last  root  fig- 
ure mentally, 
and     combin- 
ing   units    of 
like        order, 

343 

214              856 

14700 
15556 

70494 
62224 

222.5       111.25 

1642800 
16539.25 

8270.000 
8269.625 

.375 

thus :  16,  6,  and  6  are  28,  and  we  write  the  unit  figure  in  the  new 
trial  divisor ;  then  2  to  carry,  and  5  and  5  are  12,  &c.  We  annex 
2  ciphers  to  this  trial  divisor,  as  to  the  former,  and  dividing,  obtain 
5,  the  third  figure  in  the  root.  To  complete  the  second  trial  di- 
visor, after  the  manner  of  the  first,  the  correction  may  be  found  by 
annexing  .5  to  3  times  the  former  figures,  74,  and  multiplying  this 
number  by  .5.  But  as  we  have,  in  column  I,  3  times  7,  with  4 
annexed,  or  214,  we  need  only  multiply  the  last  figure,  4,  by  3, 
and  annex  .5,  making  222.5,  which  multiplied  by  .5  gives  111.25, 
the  correction  required.  Then  we  obtain  the  complete  divisor, 
16539.25,  the  product,  8269.625,  and  the  remainder,  .375,  in  the 
manner  shown  by  the  former  steps.  From  this  example  and  analysis 
we  deduce  the  following 

RULE.     I.    Point  off  the  given  number  into  periods  of  three 
figures  each,  counting  from  units'  place  toward  the  left  and  right. 

II.  Find  the  greatest  cube  that  does  not  exceed  the  left  hand 
period,  and  write  its  root  for  the  first  figure  in  the  required 
root ;  subtract  the  cube  from  the  left  hand  period,  and  to  the 
remainder  bring  down  the  next  period  for  a  dividend. 

III.  At  the  left  of  the  dividend  write  three  times  the  square 
of  the  first  figure  of  the  root,  and  annex  two  ciphers,  for  a  trial 
divisor  ;  divide  the  dividend  by  the  trial  divisor,  and  write  the 
quotient  for  a  trial  figure  w  the  root. 


324  CUBE  ROOT. 

IV.  Annex  the  trial  figure  to  three  times  the  former  figure, 
and  write  the  result  in  a  column  marked  I,  one  line  below  tlie 
trial  divisor  ;  multiply  this  term  by  the  trial  figure,  and  write 
the  product  on  the  same  line  in  a  column  marked  II ;  add  this 
term  as  a  correction  to  the  trial  divisor,  and  the  result  will  be 
the  complete  divisor. 

V.  Multiply  the  complete  divisor  by  the  trial  figure,  and 
subtract  the  product  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend. 

VI.  Add  the  square  of  the  last  figure  of  the  root,  the  last 
term  in  column  II,  and  the  complete  divisor  together  ^and  annex 
two  ciphers,  for  a  new  trial  divisor ;  with  which  obtain  an- 
other trial  figure  in  the  root. 

VII.  Multiply  the  unit  figure  of  the  last  term  in  column  I 
by  3,  and  annex  the  trial  figure  of  the  root  for  the  next  term 
of  column  I ;  multiply  this  result  by  the  trial  figure  of  the  root 
for  the  next    term  of  column  II ;  add  this  term  to  the  trial 
divisor  for  a  complete  divisor,  with  which  proceed  as  before. 

NOTES.  1.  If  at  any  time  the  product  be  greater  than  the  dividend, 
diminish  the  trial  figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the 
trial  divisor,  and  another  period  to  the  dividend ;  then  proceed  as  be- 
fore with  column  I,  annexing  both  cipher  and  trial  figure. 

EXAMPLES    FOR    PRACTICE. 

1.   What  is  the  cube  root  of  79.112  ? 

OPERATION. 

79.112  |  4.2928  +  ,  Ans. 
64. 


122       244 

4400    15112 
5044    10088 

1269     11421 

529200.   5024000 
540621   4865589 

12872    25744 

55212300  158411000 
55238044  110476088 

128768  1030144 

5526379200  47934912000 
5527409344  44219274752 

3714637248  rem. 


CUBE  ROOT.  325 

2.  What  is  the  cube  root  of  84604519  ?  Am.    439. 

3.  What  is  the  cube  root  of  2357947691  ?  Ans.  1331. 

4.  What  is  the  cube  root  of  109 63240788375?    Ans.  22215. 

5.  What  is  the  cube  root  of  270671777032189896  ? 

Ans.    646866. 

6.  What  is  the  cube  root  of  .091125  ?  Ans.    .45. 

7.  What  is  the  cube  root  of  .000529475129  ?      Ans.  .0809. 

8.  What  is  the  approximate  cube  of  .008649  ? 

Ans.   .2052  +  . 
Extract  the  cube  roots  of  the  following  numbers  :  — 


=  1.259921+ 
=.  1.442249+ 
=  1.587401+ 


=  1.709975+ 
=  1.817120+ 
—  1.912931+ 


APPLICATIONS    IN    CUBE    ROOT. 

1.  What  is  the  length  of  one  side  of  a  cistern  of  cubical 
form,  containing  1331  solid  feet?  Ans.   11  feet. 

2.  The  pedestal  of  a  certain  monument  is  a  square  block  of 
granite,  containing  373248  solid  inches ;  what  is  the  length 
of  one  of  its  sides  ?  Ans.    6  feet. 

3.  A  cubical  box  contains  474552  solid  inches  ;   what  is 
the  area  of  one  of  its  sides  ?  Ans.    42  J  sq.  ft. 

4.  How  much  paper  will  be  required  to  make  a  cubical 
box  which  shall  contain  f  £  of  a  solid  foot  ?     Ans.  f  of  a  yard. 

5.  A  man  wishes  to  make  a  bin  to  contain  125  bushels,  of 
equal  width  and  depth,  and  length  double  the  width ;  what 
must  be  its  dimensions  ?      Ans.   Width   and  depth,   51.223  + 
inches  ;  length,  102.446  +  inches. 

NOTE.  Spheres  are  to  each  other  as  the  cubes  of  their  diameters  or 
circumferences. 

6.  There  are  two  spheres  whose  solid  contents  are  to  each 
other  as  27  to  343  ;  what  is  the  ratio  of  their  diameters  ? 

ANALYSIS.  Since  spheres  are  to  each  other  as  the  cubes  of  their 
diameters,  the  diameters  will  be  to  each  other  as  the  cube  roots  of 
the  spheres ;  and  41/27  =  3,  ^/343  —  7  j  hence  the  diameters  required 
are  as  3  to  7. 

BB 


326  ARITHMETICAL   PROGRESSION. 

7.  The  diameter  of  a  sphere  containing  1  solid  foot  is  14.9 
inches ;  what  is  the  diameter  of  a  sphere  containing  2  solid 
feet  ?  Ans.    18.8  inches. 

8.  If  a  cable  measuring  4  inches  in  circumference  support 
a  weight  of  1800  pounds,  what  must  be  the  circumference  of 
a  cable  that  will  support  3515|  pounds?        Ans.   5  inches. 

ARITHMETICAL  PROGRESSION. 

4L&t3.  An  Arithmetical  Progression,  or  Series,  is  a  series 
of  numbers  increasing  or  decreasing  by  a  common  difference. 
Thus,  3,  5,  9,  11,  &c.,  is  an  arithmetical  progression  with  an 
ascending  series,  and  13,  10,  7,  4,  &c.,  is  an  arithmetical  pro- 
gression with  a  descending  series. 

4t$3.  The  Terms  of  a  series  are  the  numbers  of  which  it 
is  composed. 

434.    The  Extremes  are  the  first  and  last  terms. 

43*5.    The  Means  are  the  intermediate  terms. 

436.  The  Common  Difference  is  the  difference  between 
any  two  adjacent  terms. 

437.  There  are  Jive  parts  in  an  arithmetical  series,  any 
three  of  which  being   given,  the  other  two  may  be  found. 
They  are  as  follows :   the  first  term,  last  term,  common  differ- 
ence, number  of  terms,  and  sum  of  all  the  terms. 

CASE   I. 

438.  To   find  the  last  -term  when  the  first  term, 
common  difference,  and  number  of  terms  are  given. 

Let  2  be  the  first  term  of  an  ascending  series,  and  3  the 
common  difference;  then  the  series  will  be  written,  2,  5,  8,  11, 
14,  or  analyzed  thus :  2,  2  +  3,  2  +  3  +  3,  2  +  3  +  3  +  3, 
2  +  3  +  3  +  3  +  3. 

Here  we  see  that,  in  an  ascending  series,  we  obtain  the 
second  term  by  adding  the  common  difference  once  to  the  first 
term  ;  the  third  term,  by  adding  the  common  difference  twice 
to  the  first  term ;  and,  in  general,  we  obtain  any  term  by 


ARITHMETICAL  PROGRESSION.  327 

adding  the  common  difference  as  many  times  to  the  first  term 
as  there  are  terms  less  one. 

NOTE.  The  analysis  for  a  descending  series  would  be  similar. 
Hence, 

RULE.  Multiply  the  common  difference  by  the  number  of 
terms  less  one,  and  add  the  product  to  the  first  term,  if  the 
series  be  ascending,  and  subtract  it  if  the  series  be  descending. 

EXAMPLES. 

1.  The  first  term  of  an  ascending  series  is  4,  the  common 
difference  3,  and  the  number  of  terms  19;  what  is  the  last 
term?  Ans.   58. 

2.  What  is  the  13th  term  of  a  descending  series  whose  first 
term  is  75,  and  common  difference  5  ?  Ans.    15. 

3.  A  boy  bought  18  hens,  paying  2  cents  for  the  first,  5 
cents  for  the  second,  and  8  cents  for  the  third,  in  arithmetical 
progression ;  what  did  he  pay  for  the  last  hen  ? 

4.  What  is  the  40th  term  of  the  series  £,  -£,  1,  1^,  &c.  ? 

Ans.    10£. 

5.  A  man  travels  9  days ;   the  first  day  he  goes  20  miles, 
the  second  25  miles,  increasing  5  miles  each  day;   how  far 
does  he  travel  the  last  day  of  his  journey  ?     Ans.  60  miles. 

6.  What  is  the  amount  of  $100,  at  7  per  cent,  for  45 
years  ?  $100  +  $7  X  45  =  $415,   Ans. 

CASE    II. 

439.  To  find  the  common  difference  when  the 
extremes  and  number  of  terms  are  given. 

Referring  to  the  series,  2,  5,  8,  11,  14,  analyzed  in  438, 
we  readily  see  that,  by  subtracting  the  first  term  from  any 
term,  we  have  left  the  common  difference  taken  as  many  times 
as  there  are  terms  less  one ;  thus,  by  taking  away  2  in  the 
fifth  term,  2+3  +  3  +  3  +  3,  we  have  3  taken  4  times. 
Hence, 

RULE.  Divide  the  difference  of  the  extremes  by  the  number 
of  terms  less  one. 


328  ARITHMETICAL  PROGRESSION. 

EXAMPLES. 

1.  The  first  term  is  2,  the  last  term  is  17,  and  the  number 
of  terms  is  6  ;  what  is  the  common  difference  ?         Ans.    3. 

2.  A  man  has  seven  children,  whose  ages  are  in  arithmetical 
progression;  the  youngest  is  2  years  old,  and  the  eldest  14; 
what  is  the  common  difference  of  their  ages  ?   Ans.  2  years. 

3.  The  extremes  of  an  arithmetical  series  are  1  and  50£, 
and  the  number  of  terms  is  34 ;  what  is  the  common  difference  ? 

4.  An  invalid  commenced  to  walk  for  exercise,  increasing 
the  distance  daily  by  a  common  difference ;  the  first  day  he 
walked  3  miles,  and  the  14th  day  9£  miles;  how  many  miles 
did  he  walk  each  day  ? 

NOTE.  When  we  have  found  the  common  difference  we  may  add  it 
once,  twice,  &c.,  to  the  first  term,  and  we  have  the  series,  and  conse- 
quently the  means, 

Ans.   3,  3£,  4,  4£,  5,  5  J,  &c. 

CASE    III. 

44O.  To  find  the  number  of  terms  when  the  ex- 
tremes and  common  difference  are  given. 

Examining  the  series,  2,  5,  8,  11.  14,  analyzed  in  438  5 
we  also  see  that  after  taking  away  the  first  term  from  any 
term,  we  have  left  the  common  difference  taken  as  many 
times  as  the  number  of  terms,  less  1.  Hence, 

RULE.  Divide  the  difference  of  the  extremes  by  the  common 
difference,  and  add  1  to  the  quotient. 

EXAMPLES. 

1.  The  extremes  are  7  and  43,  and  the  common  difference 
is  4 ;  what  is  the  number  of  terms  ?  Ans.    10. 

2.  The  first  term  is  2£,  the  last  term  is  40,  and  the  common 
difference  is  7£ ;  what  is  the  number  of  terms  ?        Ans.    6. 

3.  A  laborer  agreed  to  build  a  fence  on  the  following  con- 
ditions: for  the  first  rod  he  was  to  have  6   cents,  with  an 
increase  of  4  cents  on  each  successive  rod ;  the  last  rod  came 
to  220  cents  ;  how  many  rods  did  he  build  ?    Ans.    56  rods. 


GEOMETRICAL   PROGRESSION.  329 

CASE   IV. 

441.  To  find  the  sum  of  all  the  terms  when  the 
extremes  and  number  of  terms  are  given. 

To  deduce  a  rule  for  finding  the  sum  of  all  the  terms,  we 
will  take  the  series  2,  5,  8,  11,  14,  writing  it  under  itself  in  an 
inverse  order,  and  add  each  term ;  thus, 

2+5+    8+  11  +  14  =  40,  once  the  sum. 
14  +  11+    8  +    5+    2  =  40,     «       «      « 
1C  +  16  +  1C  +  1C  +  1C  =  80,  twice  the  sum. 

Here  we  perceive  that  1C,  the  sum  of  the  extremes,  multi- 
plied by  5,  the  number  of  terms,  equals  80,  which  is  twice  the 
sum  of  the  series.  Dividing  80  by  2  ogives  40,  which  is  the 
sum  required.  Hence, 

RULE.  Multiply  the  sum  of  the  extremes  by  the  number  of 
terms,  and  divide  the  product  by  2. 

EXAMPLES. 

1.  The  extremes  are  5  and  32,  and  the  number  of  terms  12 ; 
what  is  the  sum  of  all  the  terms  ?  Am.    222. 

2.  How  many  strokes  does  a  common  clock  make  in   12 
hours  ?  Ans.    78  .strokes. 

3.  What  debt  can  be  discharged  in  a  year  by  weekly  pay- 
ments in  arithmetical  progression,  the  first  bfinjr  824,  arid  the 
last  si 224?  Ans.    $32448. 

4.  Suppose  100  apples  were  placed  in  a  line  2  yards  apart, 
and  a  basket  2  yards  from  the  first  apple ;   how  far  would  a 
boy  travel  to  gather  them  up  singly,  and  return  with  each 
separately  to  the  basket  ?  Ans.   20200  yards. 

GEOMETRICAL  PROGRESSION. 

442.  A  Geometrical  Progression  is  a  series  of  numbers 
increasing  or  decreasing  by  a  constant  multiplier. 

When  the  multiplier  is  greater  than  a  unit,  the  series  is 

BB* 


880  GEOMETRICAL  PROGRESSION. 

ascending;  thus,  2,  6,  18,  54,  162,  is  an  ascending  series,  in 
which  3  is  the  multiplier. 

When  the  multiplier  is  less  than  a  unit,  the  series  is  descend- 
ing; thus,  162,  54,  18,  6,  2,  is  a  descending  series,  in  which  -£ 
is  the  multiplier. 

443.  The  Ratio  is  the  constant  multiplier. 

444.  In   every  geometrical   progression  there  are  five 
parts  to  be  considered,  any  three  of  which  being  given,  the 
other  two  may  be  determined.    They  are  as  follows:  The  first 
term,  last  term,  ratio,  number  of  terms,  and  the  sum  of  all  the 
terms. 

The  first  and  last  terms  are  the  extremes,  and  the  interme- 
diate terms  are  the  means. 

CASE    I. 

445.  To  find  any  term,  the  first  term,  the  ratio, 
and  number  of  terms  being  given. 

The  fir.st  term  is  supposed  to  exist  independently  of  the 
ratio.  Using  the  ratio  once  as  a  factor,  we  have  the  second 
term ;  using  it  twice,  or  its  second  power,  we  have  the  third 
term ;  using  it  three  times,  or  its  third  power,  we  have  the 
fourth  term ;  and,  in  general,  the  power  of  the  ratio  in  any 
term  is  one  less  than  the  number  of  the  term.  The  ascending 
series,  2,  6,  18,  54,  may  be  analyzed  thus:  2,  2  X  3,  2  X 
3X3,  2X3X3X3. 

In  this  illustration  we  see  that 

1st  term,  2,  is  independent  of  the  ratio. 

2d  "  6  —  2  X  3  =  the  first  term  into  the  1st  power  of 
the  ratio. 

3d  term,  18  =  2  X  32  =  the  first  term  into  the  2d  power 
of  the  ratio. 

4th  term,  54  r=  2  X  3  3  =  the  first  term  into  the  3d  power 
of  the  ratio.  Hence 

RULE.  Multiply  the  fast  term  by  that  power  of  the  ratio 
denoted  by  the  number  of  terms  less  1. 


GEOMETRICAL  PROGRESSION.  331 

EXAMPLES. 

1.  The  first  term  of  a  geometrical  series  is  4,  the  ratio  is  3  ; 
what  is  the  9th  term  ?  Ans.    4  X  38  —  26244. 

2.  The  first  term  is  1024,  the  ratio  ^,  and  the  number  of 
term  8  ;  what  is  the  last  term  ?  Ans.   y1^. 

3.  A  boy  bought  9  oranges,  agreeing  to  pay  1  mill  for  the 
first  orange,  2  mills  for  the  second,  and  so  on  ;  what  did  the 
last  orange  cost  him  ?  Ans.    $2.304. 

4.  The  first  term  is  7,  the  ratio  ^-,  and  the  number  of  terms 
7  ;  what  is  the  last  term  ?  Ans.    T^§^T' 

5.  What  is  the  amount  of  $1  at  compound  interest  for  5 
years,  at  7  per  cent,  per  annum  ?  Ans.   $1.40255  +• 

NOTE.  In  the  above  example  the  first  term  is  $1,  the  ratio  is  $1.07, 
and  the  number  of  terms  is  6. 

6.  A  drover  bought  7  oxen,  agreeing  to  pay  $3  for  the  first 
ox,  $9  for  the  second,  $27  for  the  third,  and  so  on  ;  what  did 
the  last  ox  cost  him  ?  Ans.   $2187. 

CASE  n. 

446.  To  find  the  sum  of  all  the  terms,  the  ex- 
tremes and  ratio  being  given. 

If  we  take  the  series  2,  8,  32,  128,  512,  in  which  the  ratio 
is  4,  multiply  each  term  by  the  ratio,  and  add  the  terms  thus 
multiplied,  we  shall  have 

8  +  32  +  128  +  512  +  2048  =  2728  ={^  Sfe*lram 
But  2  +  8  +  32  +  128  +  512  =  682  =  j  ^&cet^ttm  of  a11 

Hence,  by  subtracting,  we  get  2048  —  2=  204G  =  {  ^^^l^ 

1  °f  *" 


Dividing  by  3,  theratio  less  one,  2046-^3  =  682  = 

The  subtraction  is  performed  by  taking  the  lower  line  or 
series  from  the  upper.  All  the  terms  cancel  except  2048  and 
2.  Taking  their  difference,  which  is  3  times  the  sum,  and  di- 
viding by  3,  the  ratio  less  one,  we  must  have  the  sum  of  all 
the  terms.  Hence 


332  PROMISCUOUS   EXAMPLES. 

RULE.  Multiply  the  greater  extreme  by  the  ratio,  subtract 
the  less  extreme  from  the  product,  and  divide  the  remainder 
by  the  ratio  less  1. 

NOTE.  Let  every  decreasing  series  be  inverted,  and  the  first  term 
called  the  last ;  then  the  ratio  will  be  greater  than  a  unit.  If  the  series 
be  infinite,  the  first  term  is  a  cipher. 

EXAMPLES. 

1.  The  first  term  is  2,  the  last  term  512,  and  the  ratio  3 ; 
what  is  the  sum  of  all  the  terms  ?  Am.   767. 

2.  The  first  term  is  4,  the  last  term  is  262144,  and  the 
ratio  is  4;  what  is  the  sum  of  the  series?       Ans.   349524. 

3.  The  first  term  of  a  descending  series  is  162,  the  last 
term  2,  and  the  ratio  £ ;  what  is  the  sum  ?  Ans.    242. 

4.  What  is  the  value  of  £,  •£§,  y^,  &c.,  to  infinity  ?  Ans.  £. 

NOTE.  In  the  following  examples  we  first  find  the  last  term  by  the 
Rule  under  Case  I. 

5.  What  yearly  debt  can  be  discharged  by  monthly  pay- 
ments, the  first  being  $2,  the  second  $6,  and  the  third  $18, 
and  so  on,  in  geometrical  progression  ?          Ans.   $531440. 

6.  If  a  grain  of  wheat  produce  7  grains,  and  these  be  sown 
the  second  year,  each  yielding  the  same  increase,  how  many 
bushels  will  be  produced  at  this  rate  in  12  years,  if  1000  grains 
make  a  pint  ?  Ans.    3604  bu.  2  pk.  1|  pt. 

7.  Six  persons  of  the  Morse  family  came  to  this  country 
200  years  ago ;  suppose  that  their  number  has  doubled  every 
20  years  since,  what  would  be  their  number  now  ? 

NOTE.  The  other  cases  in  Progression  will  be  found  in  the  Higher 
Arithmetic. 


PROMISCUOUS  EXAMPLES. 

1.  One  half  the  sum  of  two  numbers  is  800,  and  one  half  the 
difference  of  the  same  numbers  is  200  ;  what  are  the  numbers  ? 

Ans.    1000  and  600. 

2.  What  number  is  that  to  which,  if  you  add  •£  of  T\  of  itself, 
the  sum  will  be  61  ?  Ans.   55. 

3.  What  part  of  a  day  is  3  h.  21  min.  15  sec.  ?          Ans. 


PROMISCUOUS   EXAMPLES.  333 

4.  A  commission  merchant  received  70  bags  of  wheat,  each  con- 
taining 3  bu.  3  pk.  3  qt. ;  how  many  bushels  did  he  receive  ? 

5.  Four  men,  A,  B,  C,  and  D,  are  in  possession  of  $1100;   A 
has  a  certain  sum,  B  has  twice  as  much  as  A,  C  has  $300,  and  D 
has  $200  more  than  C  ;  how  many  dollars  has  A  ?       Ans.   $100. 

6.  At  a  certain  election,  3000  votes  were  cast  for  three  candi- 
dates, A,  B,  and  C  ;  B  had  200  more  votes  than  A,  and  C,  had  800 
more  than  B  ;  how  many  votes  were  cast  for  A  ?  Ans.   600. 

7.  What  part  of  17£  is  31  ?  Ans.   |f . 

8.  The  difference  between  ^  and  •£  of  a  number  is  10 ;  what  is 
the  number?  Ans.   560. 

9.  A   merchant   bought  a  hogshead  of  rum  for  $28.35  ;   how 
much  water  must  be  added  to  reduce  the  first  cost  to  35  cents  per 
gallon  ?  Ans.    18  gal. 

10.  A  and  B  traded  with  equal  sums  of  money ;  A  gained  a  sum 
equal  to  ^  of  his  stock ;  B  lost  $200,  and  then  he  had  ^  as  much  as 
A ;  how  much  was  the  original  stock  of  each  ?  Ans.   $500. 

11.  A  toner  sold  17  bushels  of  barley,  and  13  bushels  of  wheat, 
for  $31.55  ;  he  received  for  the  wheat  35  cents  a  bushel  more  than 
for  the  barley  ;  what  was  the  price  of  each  per  bushel  ? 

Ans.   Barley,  $.90;  wheat,  $1.25. 

12.  What  is  the  interval  of  time  between  March  20,  21  minutes 
past  3  o'clock,  P.  M.,  and  April  llth,  5  minutes  past  7  o'clock, 
A.  M.  ?  Ans.   21  da.  15  h.  44  min. 

13.  What  o'clock  is  it  when  the  time  from  noon  is  T9T  of  the 
time  to  midnight  ?  Ans.   5  o'clock  24  min.  P.  M. 

14.  What  is  the  least  number  of  gallons  of  wine  that  can  be 
shipped  in  hogsheads,  tierces,  or  barrels,  just  filling  the  vessels, 
without  deficit  or  excess  ?  Ans.    126  gal. 

15.  A  ferryman  has  four  boats  ;  one  will  carry  8  barrels,  another 
9,  another  15,  and  another  16  ;  what  is  the  smallest  number  of  bar- 
rels that  will  make  full  freight  for  either  one  of  the  boats  ?    Ans.  720. 

16.  A  and  B  have  the  same  income ;  A  saves  |-  of  his,  but  B, 
by  spending  $30  a  year  more  than  A,  at  the  end  of  four  years  finds 
himself  $40  in  debt ;  what  is  their  income,  and  how  much  does 
each  spend  a  year  ?  C  Income,     $160. 

Ans.  1  A  spends  $140. 
C  B  spends  $170. 

17.  If  a  load  of  plaster  weighing  1825  pounds  cost  $2.19,  how 
much  is  that  per  ton  of  2000  pounds  ?  Ans.   $2.40. 

18.  If  2|  yards  of  cloth  If  yards  wide  cost  $3.37f ,  what  will  be 
the  cost  of  36^  yards  U  yards  wide  ?  Ans.   $52.79. 

19.  I  lend  my  neighbor  $200  for  6  months  ;  how  long  ought  he 
to  lend  me  $1000  to  balance  the  favor  ?  Ans.   36  days. 

20.  Bought  railroad  stock  to  the  amount  of  $2356.80,  and  found 
that  the  sum  invested  was  40  per  cent  of  what  I  had  left ;  what 
sum  had  I  at  first  ?  Am.   $8248.80. 

21.  20  per  cent,  of  -|  of  a  number  is  what  per  cent,  of  ^  of  it  ? 

Ans.    12. 


334  PROMISCUOUS   EXAMPLES. 

22.  Bought  wheat  at  $1.50  per  bushel,  corn  at  $.75  per  bushel, 
and  barley  at  $.60  per  bushel  ;  the  wheat  cost  twice  as  much  as  the 
corn,  and  the  corn  twice  as  much  as  the  barley  ;  of  the  sum  paid, 
$243  and  J  of  the  whole  was  for  wheat,  and  $153  and  T^  of  the 
whole  was  for  the  corn  ;  how  many  bushels  of  grain  did  I  purchase  ? 

Ans.   756. 

23.  Divide  $630  among  3  persons,  so  that  the  second  shall  have 
•|  as  much  as  the  first,  and  the  third  -|  as  much  as  the  other  two  ; 
what  is  the  share  of  each  ?  C  1st,  $240. 

Ans.  32d,  $180. 
(3d,  $210. 

24.  Bought  a  hogshead  of  molasses  for   $28,  and   7   gallons 
leaked  out  ;  at  what  rate  per  gallon  must  the  remainder  be  sold  to 
gain  20  per  cent.  ?  Ans.    $.60. 

25.  20  per  cent,  of  -|  of  a  number  is  how  many  per  cent,  of  2 
times  -|  of  14  times  the  number  ?  Ans.   7^-. 

26.  B  and  C,  trading  together,  find  their  stock  to  be  worth  $3500, 
of  which  C  owns  $2100  ;  they  have  gained  40  per  cent,  on  their  first 
capital;  what  did  each  put  in  ?  A       $  B,  $1500. 


27.  If  the  ridge  of  a  building  be  8  feet  above  the  beams,  and  the 
building  be  32  feet  wide,  what  must  be  the  length  of  rafters  ? 

28.  If  12  workmen,  in  12  days,  working    12  hours  a  day,  can 
make  up  75  yards  of  cloth,  f  of  a  yard  wide,  into  articles  of  clothing  : 
how  many  yards,  1  yard  wide,  can  be  made  up  into  like  articles,  by 
10  men,  working  9  days,  8  hours  each  day  ?  Ans.   23^. 

29.  A  grocer  sells  a  farmer  100  pounds  of  sugar,  at  12  cents  a 
pound,  and  makes  a  profit  of  9  per  cent.  ;  the  former  sells  him  100 
pounds  of  beef,  at  6  cents  a  pound,  and  makes  a  profit  of  10  per 
cent.  ;  who  gains  the  most  by  the  trade,  and  how  much? 

Ans.    The  grocer  gains  $.536  -}-  most. 

30.  In   1  yr.  4  mo.    $311.50   amounted  to  $336.42,   at  simple 
interest  ;  what  was  the  rate  per  cent.  ?  Ans.    6. 

31.  Three  persons  engage  to  do  a  piece  of  work  for  $20;  A  and 
B  estimate  that  they  do  •$•  of  it,  A  and  C  that  they  do  f  of  it,  and  B 
and  C  that  they  do  f  of  it  ;  according  to  this  estimate,  what  part  of  the 
$20  should  each  man  receive  ?     Ans.  A's,  $llf  ;  B's,  $5f  ;  C's,  $2f 

32.  Paid  $375,  at  the  rate  of  2£  per  cent.,  for  insurance  on  a 
cotton  factory  and  the  machinery  ;  for  what  amount  was  the  policy 
given?  Ans.   $15000. 

33.  A  merchant  bought  goods  in  Boston  to\the  amount  of  $1000, 
and  gave  his  note,  dated  Jan.  1,  1857,  on  interest  after  90  days; 
six  months  after  the  note  Vas  given  he  paid\  $560,  and  5  months 
after  the  first  payment  he  paie^  $406  ;  what  wasYhie  Aug  23,  1859  ? 

Ans.   $63.01. 

34.  Iff  of  A's  money  be  equal  to  f  of  B's,  and  f  of  B's  be  equal 
to  -£  of  C/s,  and  f  of  C's  be  equal  to  f  of  D's,  and  D  has  $45  more 
than  C,  how  much  has  each  ?  A       $  A,  $368  ;  C,  $360  ; 

Ans'     B,  $336  ;  D,  $405. 


Tffep 

PROMISCUOUS   EXJBTLHS.      w  W 

'KSI.TY 

35.  A  owed  B  $900,  to  be  paid  in  S^barsy^  but  aylafi.  expiration 
of  9  months  A  agreed  to  pay  $300  if  B  V^Ejdk^oiig  enough  for 
the  balance  to  compensate  for  the  advancVj^hcw  long  should  B 
wait  after  the  expiration  of  the  3  years  ?  Arts,    lo^  mo. 

36.  A  certain  clerk  receives  $800  a  year ;  his  expenses  equal  ^ 
of  what  he  saves ;  how  much  of  his  salary  does  he  save  yearly  ? 

37.  A  merchant  sold  cloth  at  $1  per  yard,  and  made  10  per  cent, 
profit ;  what  would  have  been  his  gain  or  loss  had  he  sold  it  at  $.87-^ 
per  yard  ?  Ans.   Loss,  3|  per  cent. 

21— 9 

38.  What  is  the  cube  of  -~  Ans.  fj. 

11  63 

39.  What  is  the  cube  root  of  — —  Ans.  f . 

40.  A  miller  is  required  to  grind  100  bushels  of  provender  worth 
50  cents  a  bushel,  from  oats  worth  20  cents,  corn  worth  35  cents, 
rye  worth  60  cents,  and  wheat  worth  70  cents  per  bushel;    how 
many  bushels  of  each  must  he  take  ? 

41.  A  man  owesl$6480  to  his  creditors;  his  debts  are  in  arith- 
metical progression,!  the  least  being  $40,  and  the  greatest  $500 ; 
required  the  numberlof  creditors  aad  the  common  difference  between 
the  debts.  A       5  24  creditors. 

s'  I  $20  difference. 

42.  Two  ships  sail  from  the  same  port ;  one  goes  due  north  128 
miles,  and  the  other  due  east  72  miles ;  how  far  are  the  ships  from 
each  other  ?  Ans.    146.86  -|-  miles. 

43.  If  10  pounds  of  cheese  are  equal  in  value  to  7  pounds  of 
butter,  and  11  pounds  of  butter  to  2  bushels  of  corn,  and  14  bushels 
of  corn  to  8  bushels  of  rye,  and  4  bushels  of  rye  to  1  cord  of  wood ; 
how  many  pounds  of  cheese  are  equal  in  value* to  10  cords  of  wood? 

Ans.    550. 

44.  A  and  B  trade  until  they  gain  6  per  cent,  on  their  stock ; 
then  f  of  A's  gain  is  $18 ;  now,  if  A's  stock  is  to  B's  as  -f  to  ^,  how 
much  did  each  gain,  and  wrhat  was  the  original  stock  of  each  ? 

A       5  A's  gain,  $45  ;      stock,  $750. 
S"  I  B's      "     $37.50 ;    "       $625. 

45.  If  20  men,  in  21  days,  by  working  10  hours  a  day,  can  dig  a 
trench  30  ft.  long,  15  ft.  wide,  and  12  ft.  deep,  when  the  ground  is 
called  3  degrees  of  hardness,  how  many  men,  in  25  days,  by  work- 
ing 8  hours  a  day,  can  dig  another  trench  45  ft.  long,  16  ft.  wide, 
and   18  ft.  deep,  when  the  ground  is  estimated  at  5  degrees  of 
hardness  ?  Ans.   42. 

46.  Wishing  to  know  the  height  of  a  certain  steeple,  I  measured 
the  shadow  of  the  same  on  a  horizontal  plane,  27^  feet ;  I  then 
erected  a  10  feet  pole  on  the  same  plane,  and  it  cast  a  shadow  of  2-| 
feet ;  what  was  the  height  of  the  steeple  ?  Ans.    1031  ft. 

47.  A  can  do  a  piece  of  work  in  3  days,  B  can  do  3  times  as 
much  in  8  days,  and  C  5  times  as  much  in  12  days ;  in  what  time 
can  they  all  do  the  first  piece  of  work  ?  Ans.   f  da. 


336  PROMISCUOUS   EXAMPLES. 

48.  How  many  building  lots,  each  75  feet  by  125  feet,  can  be 
made  from  1  A.  1  R.  6  P.  18£  sq.  yd.  ?  Ans.    6. 

49.  A  man  bought  a  house,  and  as/eed  to  pay  for  it  $1  on  the 
first  day  of  January,  $2  on  the  first/day  of  February,  $4  on  the 
first    day  March,  and  so  on,  in  geometrical   progression,  through 
the  year  ;  what  will  be  the  cost  of  the  house,  and  what  the  average 
time  of  payment  ?  A      \  $4095- 

's'  I  Average  time,  Nov.  1. 

50.  A  man  sold  a  rectangular  piece  of  ground,  measuring  44 
chains  32  links  long  by  36  chains  wide;    how  many  acres  did  it 
contain  ?  Ans.    159  A.  2  R.  8.32  P. 

51.  What  number  is  that  which  being  increased  by  its  half,  its 
third,  and  18  more,  will  be  doubled  ?  Ans.    108^. 

52.  A  merchant  has  200  Ib.  of  tea,  worth  $.62^  per  pound,  which 
he  will  sell  at  $.56  per  pound,  provided  the  purchaser  will  pay  in 
coffee  at  22  cents,  which  is  worth  25  cents  per  pound ;  does  the 
merchant  gain  or  lose  by  the  sale  of  the  tea,  and  how  much  per 
cent.  ?  Ans.   Lost  1H  per  cent. 

53.  A  man  owes  a  debt  to  be  paid  in  4  equal  installments  at  4, 
9,  12,  and  20  months,  respectively;  discount  being  allowed  at  5  per 
cent.,  he  finds  that  $750  ready  money  will  pay  the  debt ;  how  much 
did  he  owe?  Ans.   $784.74+. 

54.  A  and  B  traded  upon  equal  capitals  ;  A  gained  a  sum  equal 
to  -|  of  his  capital,  and  B  a  sum  equal  to  -|l  of  his ;  B's  gain  was 
$500  less  than  A's  ;  what  was  the  capital  of  each  ?     Ans.  $4000. 

55.  I  purchase  goods  in  bills  as  follows:    June  4,  1859,  $240.75; 
Aug.  9,  1859,  $|137.25;    Auk  29,   1859,  $65.64;    Sept.  4,  1859, 
$230.36;    Nov.   12,   1859,  $$3.     If  the  merchant  agree  to  allow 
credit  of  6  mo.  qjn  each  bill,  wnen  may  I  settle  by  paying  the  whole 
amount?  Ans.   Feb.  1,  1860. 

56.  A  young  man  inherited  a  fortune,  ^  of  which  he  spent  in  3 
months,  and  $  of  the  remainder  in  10  months,  when  he  had  only 
$2524  left;  how  much  had  he  at  first?  Ans.  $5889.33  +. 

57.  A  man  bought  a  piece  of  land  for  $3000,  agreeing  to  pay  7 
per  cent  interest,  and  to  pay  principaV  and  interest  in  5  equal  an- 
nual installments  ;  how  much  was  the'  annual  payment  ? 

Ans.   $731.67-}-. 

58.  A  man  held  three  notes,  the  first  for  $600,  due  July  7,  1859 ; 
the  second  for  $530,  due  Oct.  4,  1859 ;  and  the  third  for  $400,  due 
Feb.   20,   1860;   he  'made  an  equitable  exchange  of  these  with  a 
speculator  for  two  o,ther  notes,  one  of  which  was  for  $730,  due  Nov. 
15,  1859;  what  was  the  face  of  the  other,  and  when  due? 

A       J  Face,  $800. 

s'  I  Due  Aug.  29,  1859. 

59.  A  room  is  27  ft.  6  in.  by  22  ft.  6  in.,  and  10  ft.  3  in.  above  the 
base-board ;    in  said  room  are  2  doors,  each  8  ft.  by  4  ft.  4  in. ;  2 
windows,  each  6  ft.  by  3  ft.  4  in.,  and  a  fireplace  of  4  ft.  by  4  ft.  6 
in. :  what  will  it  cost  to  plaster  the  room,  at  18  cents  per  square 
yard?  Ans.   $30.328$. 


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